Exponents and Radicals : Exercise (2.4) - Solution

1. Simplify the following.
   
   (a) $3 \sqrt{5}+7 \sqrt{5}$
   
   
   
   

   $ \begin{aligned} &3\sqrt{5}+7\sqrt{5}\\\\ =&10\sqrt{5}\end{aligned}$

   
   
   (b) $\sqrt{75}-\sqrt{12}$
   
   
   
   

   $\begin{aligned} &\sqrt{{75}}-\sqrt{{12}}\\\\ =&\sqrt{{25\times 3}}-\sqrt{{4\times 3}}\\\\ =&5\sqrt{3}-2\sqrt{3}\\\\ =&3\sqrt{3} \end{aligned}$

   
   
   (c) $3 \cdot 3 \sqrt{3} \cdot 3 \sqrt{27}$
   
   
   
   

   $\begin{aligned} &3\cdot 3\sqrt{3}\cdot 3\sqrt{{27}}\\\\ =&27\cdot \sqrt{3}\cdot \sqrt{{27}}\\\\ =&27\cdot \sqrt{3}\cdot \sqrt{{9\times 3}}\\\\ =&27\cdot \sqrt{3}\cdot 3\sqrt{3}\\\\ =&81\cdot \sqrt{3}\cdot \sqrt{3}\\\\ =&81\cdot 3\\\\ =&243 \end{aligned}$

   
   
   (d) $2 \sqrt{5} \cdot 3 \sqrt{2}$
   
   
   
   

   $\begin{aligned} &2 \sqrt{5} \cdot 3 \sqrt{2}\\\\ =&6\sqrt{10} \end{aligned}$

   
   
   (e) $(4-\sqrt{3})^{2}$
   
   
   
   

   $\begin{aligned} &(4-\sqrt{3})^2\\\\ =&16-2\times 4\times \sqrt{3}+3\\\\ =&19-8\sqrt{3} \end{aligned}$

   
   
   (f) $(\sqrt{3}+2 \sqrt{2})(\sqrt{3}+\sqrt{2})$
   
   
   
   

   $\begin{aligned} &\left( {\sqrt{3}+2\sqrt{2}} \right)\left( {\sqrt{3}+\sqrt{2}} \right)\\\\ =&\sqrt{3}\left( {\sqrt{3}+2\sqrt{2}} \right)+\sqrt{2}\left( {\sqrt{3}+2\sqrt{2}} \right)\\\\ =&3+2\sqrt{6}+\sqrt{6}+4\\\\ =&7+3\sqrt{6} \end{aligned}$

   
   
   (g) $(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})(\sqrt{x}+1)(\sqrt{x}-1)$
   
   
   
   

   $\begin{aligned} &\left( {\sqrt{7}-\sqrt{6}} \right)\left( {\sqrt{7}+\sqrt{6}} \right)\left( {\sqrt{x}+1} \right)\left( {\sqrt{x}-1} \right)\\\\ =&\left( {7-6} \right)\left( {x-1} \right)\\\\ =&x-1 \end{aligned}$

   
   
   (h) $\sqrt{75}-\dfrac{3}{4} \sqrt{48}-5 \sqrt{12}$
   
   
   
   

   $\begin{aligned} &\sqrt{{75}}-\frac{3}{4}\sqrt{{48}}-5\sqrt{{12}}\\\\ =&\sqrt{{25\times 3}}-\frac{3}{4}\sqrt{{16\times 3}}-5\sqrt{{4\times 3}}\\\\ =&5\sqrt{3}-\frac{3}{4}\times 4\sqrt{3}-5\times 2\sqrt{3}\\\\ =&5\sqrt{3}-3\sqrt{3}-10\sqrt{3}\\\\ =&-8\sqrt{3} \end{aligned}$

   
   
   (i) $\sqrt{2 x^{2}}+5 \sqrt{32 x^{2}}-2 \sqrt{98 x^{2}}$
   
   
   
   

   $\begin{aligned} &\sqrt{{2{{x}^{2}}}}+5\sqrt{{32{{x}^{2}}}}-2\sqrt{{98{{x}^{2}}}}\\\\ =&\sqrt{{2{{x}^{2}}}}+5\sqrt{{16\times 2{{x}^{2}}}}-2\sqrt{{49\times 2{{x}^{2}}}}\\\\ =&\left( {\sqrt{2}x} \right)+\left( {5\times 4\sqrt{2}x} \right)-\left( {2\times 7\sqrt{2}x} \right)\\\\ =&\sqrt{2}x+20\sqrt{2}x-14\sqrt{2}x\\\\ =&7\sqrt{2}x \end{aligned}$

   
   
   (j) $\sqrt{20 a^{3}}+a \sqrt{5 a}+\sqrt{80 a^{3}}$
   
   
   
   

   $\begin{aligned} &\sqrt{{20{{a}^{3}}}}+a\sqrt{{5a}}+\sqrt{{80{{a}^{3}}}}\\\\ =&\sqrt{{4\times 5{{a}^{3}}}}+a\sqrt{{5a}}+\sqrt{{16\times 5{{a}^{3}}}}\\\\ =&2a\sqrt{{5a}}+a\sqrt{{5a}}+4a\sqrt{{5a}}\\\\ =&7a\sqrt{{5a}} \end{aligned}$

   
   
   
2. Rationalise the denominators and simplify.
   
   (a) $\dfrac{2}{\sqrt{5}}$
   
   
   
   

   $\begin{aligned} &\dfrac{2}{\sqrt{5}}\\\\ =&\dfrac{2}{\sqrt{5}}\times\dfrac{\sqrt{5}}{\sqrt{5}}\\\\ =&\dfrac{2\sqrt{5}}{5} \end{aligned}$

   
   
   (b) $\dfrac{5}{2+\sqrt{3}}$
   
   
   
   

   $\begin{aligned} &\dfrac{5}{{2+\sqrt{3}}}\\\\ =&\dfrac{{5\left( {2-\sqrt{3}} \right)}}{{\left( {2+\sqrt{3}} \right)\left( {2-\sqrt{3}} \right)}}\\\\ =&\dfrac{{5\left( {2-\sqrt{3}} \right)}}{{4-3}}\\\\ =&5\left( {2-\sqrt{3}} \right) \end{aligned}$

   
   
   (c) $\dfrac{12}{\sqrt{5}-\sqrt{3}}$
   
   
   
   

   $\begin{aligned} &\dfrac{{12}}{{\sqrt{5}-\sqrt{3}}}\\\\ =&\dfrac{{12\left( {\sqrt{5}+\sqrt{3}} \right)}}{{\left( {\sqrt{5}-\sqrt{3}} \right)\left( {\sqrt{5}+\sqrt{3}} \right)}}\\\\ =&\dfrac{{12\left( {\sqrt{5}+\sqrt{3}} \right)}}{{5-3}}\\\\ =&\dfrac{{12\left( {\sqrt{5}+\sqrt{3}} \right)}}{2}\\\\ =&6\left( {\sqrt{5}+\sqrt{3}} \right) \end{aligned}$

   
   
   (d) $\dfrac{\sqrt{2}+1}{2 \sqrt{2}-1}$
   
   
   
   

   $\begin{aligned} &\dfrac{{\sqrt{2}+1}}{{2\sqrt{2}-1}}\\\\ =&\dfrac{{\left( {\sqrt{2}+1} \right)\left( {2\sqrt{2}+1} \right)}}{{\left( {2\sqrt{2}-1} \right)\left( {2\sqrt{2}+1} \right)}}\\\\ =&\dfrac{{\sqrt{2}\left( {2\sqrt{2}+1} \right)+1\left( {2\sqrt{2}+1} \right)}}{{8-1}}\\\\ =&\dfrac{{4+\sqrt{2}+2\sqrt{2}+1}}{7}\\\\ =&\dfrac{{5+3\sqrt{2}}}{7} \end{aligned}$

   
   
   (e) $\dfrac{\sqrt{7}+3 \sqrt{2}}{\sqrt{7}-\sqrt{2}}$
   
   
   
   

   $\begin{aligned} &\dfrac{{\sqrt{7}+3\sqrt{2}}}{{\sqrt{7}-\sqrt{2}}}\\\\ =&\dfrac{{\left( {\sqrt{7}+3\sqrt{2}} \right)\left( {\sqrt{7}+\sqrt{2}} \right)}}{{\left( {\sqrt{7}-\sqrt{2}} \right)\left( {\sqrt{7}+\sqrt{2}} \right)}}\\\\ =&\dfrac{{\sqrt{7}\left( {\sqrt{7}+\sqrt{2}} \right)+3\sqrt{2}\left( {\sqrt{7}+\sqrt{2}} \right)}}{{7-2}}\\\\ =&\dfrac{{7+\sqrt{{14}}+3\sqrt{{14}}+6}}{5}\\\\ =&\dfrac{{13+4\sqrt{{14}}}}{5} \end{aligned}$

   
   
   (f) $\dfrac{\sqrt{17}-\sqrt{11}}{\sqrt{17}+\sqrt{11}}$
   
   
   
   

   $\begin{aligned} &\dfrac{{\sqrt{{17}}-\sqrt{{11}}}}{{\sqrt{{17}}+\sqrt{{11}}}}\\\\ =&\dfrac{{\left( {\sqrt{{17}}-\sqrt{{11}}} \right)\left( {\sqrt{{17}}-\sqrt{{11}}} \right)}}{{\left( {\sqrt{{17}}+\sqrt{{11}}} \right)\left( {\sqrt{{17}}-\sqrt{{11}}} \right)}}\\\\ =&\dfrac{{17-2\sqrt{{17}}\sqrt{{11}}+11}}{{17-11}}\\\\ =&\dfrac{{28-2\sqrt{{187}}}}{6}\\\\ =&\dfrac{{2\left( {14-\sqrt{{187}}} \right)}}{6}\\\\ =&\dfrac{{14-\sqrt{{187}}}}{3} \end{aligned}$

   
   
   (g) $\dfrac{1}{2 \sqrt{2}-\sqrt{3}}$
   
   
   
   

   $\begin{aligned} &\dfrac{1}{{2\sqrt{2}-\sqrt{3}}}\\\\ =&\dfrac{{1\cdot \left( {2\sqrt{2}+\sqrt{3}} \right)}}{{\left( {2\sqrt{2}-\sqrt{3}} \right)\left( {2\sqrt{2}+\sqrt{3}} \right)}}\\\\ =&\dfrac{{2\sqrt{2}+\sqrt{3}}}{{8-3}}\\\\ =&\dfrac{{2\sqrt{2}+\sqrt{3}}}{5} \end{aligned}$

   
   
   (h) $\dfrac{\sqrt{6}+1}{3-\sqrt{5}}$
   
   
   
   

   $\begin{aligned} &\dfrac{{\sqrt{6}+1}}{{3-\sqrt{5}}}\\\\ =&\dfrac{{\left( {\sqrt{6}+1} \right)\left( {3+\sqrt{5}} \right)}}{{\left( {3-\sqrt{5}} \right)\left( {3+\sqrt{5}} \right)}}\\\\ =&\dfrac{{\sqrt{6}\left( {3+\sqrt{5}} \right)+1\left( {3+\sqrt{5}} \right)}}{{9-5}}\\\\ =&\dfrac{{3\sqrt{6}+\sqrt{{30}}+3+\sqrt{5}}}{4}\\\\ =&\dfrac{{3+\sqrt{5}+3\sqrt{6}+\sqrt{{30}}}}{4} \end{aligned}$

   
   
   
3. Write as a single fraction.
   
   (a) $\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}$
   
   
   
   

   $\begin{aligned} & \dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1} \\\\ =& \dfrac{1}{\sqrt{3}+1} \times \dfrac{\sqrt{3}-1}{\sqrt{3}-1}+\dfrac{1}{\sqrt{3}-1} \times \dfrac{\sqrt{3}+1}{\sqrt{3}+1} \\\\ =& \dfrac{\sqrt{3}-1}{(\sqrt{3})^{2}-1^{2}}+\dfrac{\sqrt{3}+1}{(\sqrt{3})^{2}-1^{2}} \\\\ =& \dfrac{\sqrt{3}-1}{3-1}+\dfrac{\sqrt{3}+1}{3-1} \\\\ =& \dfrac{\sqrt{3}-1+\sqrt{3}+1}{2} \\\\ =& \dfrac{2 \sqrt{3}}{2} \\\\ =& \sqrt{3} \end{aligned}$

   
   
   (b) $\dfrac{2}{\sqrt{7}+\sqrt{2}}+\dfrac{1}{\sqrt{7}-\sqrt{2}}$
   
   
   
   

   $\begin{aligned} & \dfrac{2}{\sqrt{7}+\sqrt{2}}+\dfrac{1}{\sqrt{7}-\sqrt{2}} \\\\ =& \dfrac{2}{\sqrt{7}+\sqrt{2}} \times \dfrac{\sqrt{7}-\sqrt{2}}{\sqrt{7}-\sqrt{2}}+\dfrac{1}{\sqrt{7}-\sqrt{2}} \times \dfrac{\sqrt{7}+\sqrt{2}}{\sqrt{7}+\sqrt{2}} \\\\ =& \dfrac{2(\sqrt{7}-\sqrt{2})}{(\sqrt{7})^{2}-(\sqrt{2})^{2}}+\dfrac{\sqrt{7}+\sqrt{2}}{(\sqrt{7})^{2}-(\sqrt{2})^{2}} \\\\ =& \dfrac{2(\sqrt{7}-\sqrt{2})}{7-2}+\dfrac{\sqrt{7}+\sqrt{2}}{7-2} \\\\ =& \dfrac{2 \sqrt{7}-2 \sqrt{2}+\sqrt{7}+\sqrt{2}}{5} \\\\ =& \dfrac{3 \sqrt{7}-\sqrt{2}}{5} \end{aligned}$

   
   
   (c) $\dfrac{1}{3+\sqrt{3}}+\dfrac{1}{\sqrt{3}-3}+\dfrac{1}{\sqrt{3}}$
   
   
   
   

   $\begin{aligned} & \dfrac{1}{3+\sqrt{3}}+\dfrac{1}{\sqrt{3}-3}+\dfrac{1}{\sqrt{3}} \\\\ =& \dfrac{3-3 \sqrt{3}}{\sqrt{3}(3+\sqrt{3})(\sqrt{3}-3)}+\dfrac{3 \sqrt{3}+3}{\sqrt{3}(3+\sqrt{3})(\sqrt{3}-3)}+\dfrac{-6}{\sqrt{3}(3+\sqrt{3})(\sqrt{3}-3)} \\\\ =& \dfrac{3-3 \sqrt{3}+3 \sqrt{3}+3-6}{\sqrt{3}(3+\sqrt{3})(\sqrt{3}-3)} \\\\ =& \dfrac{0}{\sqrt{3}(3+\sqrt{3})(\sqrt{3}-3)} \\\\ =& 0 \end{aligned}$

   
   
   (d) $\dfrac{7+\sqrt{5}}{7-\sqrt{5}}+\dfrac{\sqrt{11}-3}{\sqrt{11}+3}$
   
   
   
   

   $\begin{aligned} & \dfrac{7+\sqrt{5}}{7-\sqrt{5}}+\dfrac{\sqrt{11}-3}{\sqrt{11}+3} \\\\ =& \dfrac{(7+\sqrt{5})(7+\sqrt{5})}{(7-\sqrt{5})(7+\sqrt{5})}+\dfrac{(\sqrt{11}-3)(\sqrt{11}-3)}{(\sqrt{11}+3)(\sqrt{11}-3)} \\\\ =& \dfrac{54+14 \sqrt{5}}{49-5}+\dfrac{20-6 \sqrt{11}}{11-9} \\\\ =& \dfrac{54+14 \sqrt{5}}{44}+\dfrac{20-6 \sqrt{11}}{2} \\\\ =& \dfrac{27+7 \sqrt{5}}{22}+10-3 \sqrt{11} \\\\ =& \dfrac{27+7 \sqrt{5}+220-66 \sqrt{11}}{22} \\\\ =& \dfrac{247+7 \sqrt{5}-66 \sqrt{11}}{22} \end{aligned}$

   
   
   (e) $\dfrac{3+2 \sqrt{2}}{(\sqrt{3}-1)^{2}}$
   
   
   
   

   $\begin{aligned} & \dfrac{3+2 \sqrt{2}}{(\sqrt{3}-1)^{2}} \\\\ =& \dfrac{3+2 \sqrt{2}}{(\sqrt{3})^{2}-2 \sqrt{3+1}} \\\\ =& \dfrac{3+2 \sqrt{2}}{4-2 \sqrt{3}} \times \dfrac{4+2 \sqrt{3}}{4+2 \sqrt{3}} \\\\ =& \dfrac{(3+2 \sqrt{2})(4+2 \sqrt{3})}{4^{2}-(2 \sqrt{3})^{2}} \\\\ =& \dfrac{12+6 \sqrt{3}+8 \sqrt{2}+4 \sqrt{6}}{16-12} \\\\ =& \dfrac{12+6 \sqrt{3}+8 \sqrt{2}+4 \sqrt{6}}{4} \\\\ =& \dfrac{6+3 \sqrt{3}+4 \sqrt{2}+2 \sqrt{6}}{2} \end{aligned}$

   
   
   (f) $\sqrt{\dfrac{x+1}{x-1}}+\sqrt{\dfrac{x-1}{x+1}}-\sqrt{\dfrac{1}{x^{2}-1}}$
   
   
   
   

   $\begin{aligned} & \sqrt{\dfrac{x+1}{x-1}}+\sqrt{\dfrac{x-1}{x+1}}-\sqrt{\dfrac{1}{x^{2}-1}} \\\\ =& \dfrac{\sqrt{x+1}}{\sqrt{x-1}}+\dfrac{\sqrt{x-1}}{\sqrt{x+1}}-\dfrac{1}{\sqrt{x^{2}-1}} \\\\ =&\left(\dfrac{\sqrt{x+1}}{\sqrt{x-1}} \times \dfrac{\sqrt{x-1}}{\sqrt{x-1}}\right)+\left(\dfrac{\sqrt{x-1}}{\sqrt{x+1}} \times \dfrac{\sqrt{x+1}}{\sqrt{x+1}}\right)-\left(\dfrac{1}{\sqrt{x^{2}-1}} \times \dfrac{\sqrt{x^{2}-1}}{\sqrt{x^{2}-1}}\right) \\\\ =& \dfrac{\sqrt{x^{2}-1}}{x-1}+\dfrac{\sqrt{x^{2}-1}}{x+1}-\dfrac{\sqrt{x^{2}-1}}{x^{2}-1} \\\\ =& \dfrac{(x+1) \sqrt{x^{2}-1}}{(x-1)(x+1)}+\dfrac{(x-1) \sqrt{x^{2}-1}}{(x+1)(x-1)}-\dfrac{\sqrt{x^{2}-1}}{x^{2}-1} \\\\ =&\left(\dfrac{x+1+x-1-1}{x^{2}-1}\right) \sqrt{x^{2}-1} \\\\ =& \dfrac{(2 x+1) \sqrt{x^{2}-1}}{x^{2}-1} \end{aligned}$

   
   
   (g) $\sqrt{\dfrac{\sqrt[5]{32}+\sqrt{4}}{2^{-2}-2^{-3}}}$
   
   
   
   

   $\begin{aligned} & \sqrt{\dfrac{\sqrt[5]{32}+\sqrt{4}}{2^{-2}-2^{-3}}} \\\\ =& \sqrt{\dfrac{\sqrt[5]{2^{5}}+\sqrt{4}}{\dfrac{1}{2^{2}}-\dfrac{1}{2^{3}}}} \\\\ =& \sqrt{\dfrac{2+2}{\dfrac{1}{4}-\dfrac{1}{8}}} \\\\ =& \sqrt{\dfrac{4}{1 / 8}} \\\\ =& \sqrt{4 \times 8} \\\\ =& \sqrt{32} \\\\ =& 4 \sqrt{2} \end{aligned}$

   
   
   

Grade 10 - Exponents and Radicals (Multiple Choice Questions)

ဖြေကြည့်ပါ...၊ ရမှတ်နဲ့ အဖြေမှန်ကို ဖော်ပြပေးပါလိမ့်မယ်... ရှင်းလင်းချက် မပါဝင်ပါ။

Exponents and Radicals : Exercise (2.3) - Solutions

1.         Write the following in radical form.

$\begin{array}{ll} \text{(a)}\quad (5)^{^{\tfrac{1}{2}}} & \text{(b)}\quad (-9)^{^{\tfrac{1}{3}}} \\\\ \text{(c)}\quad (2)^{^{-\frac{1}{2}}} & \text{(d)}\quad \left(-\displaystyle\frac{3}{4}\right)^{^{\tfrac{2}{5}}}\\\\ \text{(e)}\quad \left(\displaystyle\frac{2}{7}\right)^{^{\tfrac{5}{2}}} \end{array}$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\ \ {{\left( 5 \right)}^{{\tfrac{{1\,}}{{2\,}}}}}=\sqrt{5}\\\\ \text{(b)}\ \ {{\left( {-9} \right)}^{{\tfrac{1}{3}}}}=\sqrt[3]{{-9}}\\\\ \text{(c)}\ \ {{\left( 2 \right)}^{{-\tfrac{{1\,}}{{2\,}}}}}=\displaystyle \frac{1}{{{{{\left( 2 \right)}}^{{\tfrac{{1\,}}{{2\,}}}}}}}=\displaystyle \frac{1}{{\sqrt{2}}}\\\\ \text{(d)}\ \ {{\left( {-\displaystyle \frac{3}{4}} \right)}^{{\tfrac{2}{5}}}}=\sqrt[5]{{{{{\left( {-\displaystyle \frac{3}{4}} \right)}}^{2}}}}=\sqrt[5]{{\displaystyle \frac{9}{{16}}}}\\\\ \text{(e)}\ \ {{\left( {\displaystyle \frac{2}{7}} \right)}^{{\tfrac{5}{2}}}}=\sqrt{{{{{\left( {\displaystyle \frac{2}{7}} \right)}}^{5}}}}=\sqrt{{\displaystyle \frac{{32}}{{16807}}}}\\ \end{array}$

2.         Write the following in fractional exponent form.

$\begin{array}{ll} \text{(a)}\quad \sqrt[6]{c^{^{5}}} & \text{(b)}\quad \sqrt[3]{-2} \\\\ \text{(c)}\quad \sqrt[5]{a^{^{4}} \sqrt[3]{b^{^{5}}}} & \text{(d)}\quad \sqrt[4]{\left(\displaystyle\frac{3}{7}\right)^{^{3}}} \end{array}$

Show/Hide Solution

$\begin{array}{l} (\text{a})\ \ \sqrt[6]{{{{c}^{5}}}}={{c}^{{\frac{5}{6}}}}\\\\ (\text{b})\ \ \sqrt[3]{{-2}}={{(-2)}^{{\frac{1}{3}}}}\\\\ (\text{c})\ \ \sqrt[5]{{{{a}^{4}}\sqrt[3]{{{{b}^{5}}}}}}=\sqrt[5]{{{{a}^{4}}{{b}^{{\frac{5}{3}}}}}}={{\left( {{{a}^{4}}{{b}^{{\frac{5}{3}}}}} \right)}^{{\frac{1}{5}}}}={{a}^{{\frac{4}{5}}}}{{b}^{{\frac{1}{3}}}}\\\\ (\text{d})\ \ \sqrt[4]{{{{{\left( {\displaystyle \frac{3}{7}} \right)}}^{3}}}}={{\left( {\displaystyle \frac{3}{7}} \right)}^{{\frac{3}{4}}}} \end{array}$

3.         Change the expression with the same radical and simplify the radicands.

$\begin{array}{ll} \text{(a)}\quad 6 \sqrt{2} \quad& \text{(b)}\quad 3 a \sqrt[3]{x} \\\\ \text{(c)}\quad 2 \sqrt[5]{2} \quad& \text{(d)}\quad \sqrt[4]{\displaystyle\frac{1}{2}}\\\\ \text{(e)}\quad 3 \sqrt{x^{^{3}}} \end{array}$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\quad 6 \sqrt{2}=\sqrt{6^{2} \cdot 2}=\sqrt{72}\\\\ \text{(b)}\quad 3 a \sqrt[3]{x}=\sqrt[3]{3^{3} \cdot a^{3} x}=\sqrt[3]{27 a^{3} x}\\\\ \text{(c)}\quad 2 \sqrt[5]{2}=\sqrt[5]{2^{5} \cdot 2}=\sqrt[5]{64}\\\\ \text{(d)}\quad 3 \sqrt[4]{\displaystyle\frac{1}{2}}=\sqrt[4]{3^{4} \cdot \displaystyle\frac{1}{2}}=\sqrt[4]{\displaystyle\frac{81}{2}}\\\\ \text{(e)}\quad 3 \sqrt{x^{3}}=\sqrt{3^{2} \cdot x^{3}}=\sqrt{9 x^{3}} \end{array}$

4.         Simplify.

$\begin{array}{ll} \text{(a)}\quad \sqrt{32} & \text{(b)}\quad \sqrt[5]{-32} \\\\ \text{(c)}\quad \sqrt[4]{\displaystyle\frac{81 x^{^{16}}}{16 y^{^{4}}}} & \text{(d)}\quad \sqrt[3]{\displaystyle\frac{81 x^{^{2}}}{4 y}} \\\\ \text{(e)}\ \displaystyle\frac{9^{^{\tfrac{1}{2}}}}{\sqrt[3]{27}}& \text{(f)}\quad \sqrt{\displaystyle\frac{2}{3}} \cdot \sqrt{\displaystyle\frac{75}{98}}\\\\ \text{(g)}\quad \sqrt[3]{\displaystyle\frac{-216}{8 \times 10^{^{3}}}} & \text{(h)}\quad \sqrt[n]{\displaystyle\frac{32}{2^{^{5+n}}}} \end{array}$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\quad \sqrt{32}=\sqrt{16 \cdot 2}=4 \sqrt{2}\\\\ \text{(b)}\quad \sqrt[5]{-32}=\sqrt[5]{(-2)^{5}}=-2\\\\ \text{(c)}\quad \sqrt[4]{\displaystyle\frac{81 x^{16}}{16 y^{4}}}=\sqrt[4]{\displaystyle\frac{3^{4}\left(x^{4}\right)^{4}}{2^{4} \cdot y^{4}}}=\displaystyle\frac{3 x^{4}}{2 y}\\\\ \text{(d)}\quad \sqrt[3]{\displaystyle\frac{81 x^{2}}{4 y}}=\sqrt[3]{\displaystyle\frac{3^{3} \cdot 3 x^{2}}{4 y}}=3 \sqrt[3]{\displaystyle\frac{3 x^{2}}{4 y}}\\\\ \text{(e)}\quad \displaystyle\frac{9^{\displaystyle\frac{1}{2}}}{\sqrt[3]{27}}=\displaystyle\frac{\left(3^{2}\right)^{\frac{1}{2}}}{\sqrt[3]{3^{3}}}=\displaystyle\frac{3}{3}=1\\\\ \text{(f)}\quad \sqrt{\displaystyle\frac{2}{3}} \cdot \sqrt{\displaystyle\frac{75}{98}}=\sqrt{\displaystyle\frac{2}{3} \times \displaystyle\frac{75}{98}}=\sqrt{\displaystyle\frac{25}{49}}=\displaystyle\frac{5}{7}\\\\ \text{(g)}\quad \sqrt[3]{\displaystyle\frac{-216}{8 \times 10^{3}}}=\sqrt[3]{\displaystyle\frac{(-6)^{3}}{2^{3} \times 10^{3}}}=\displaystyle\frac{-6}{2 \times 10}=-\displaystyle\frac{3}{10}\\\\ \text{(h)}\quad \sqrt[4]{\displaystyle\frac{32}{2^{5+n}}}=\sqrt[n]{\displaystyle\frac{2^{5}}{2^{5+n}}}=\sqrt[n]{\displaystyle\frac{1}{2^{n}}}=\displaystyle\frac{1}{2} \end{array}$

5.         Rationalize the denominators.

$\begin{array}{ll} \text{(a)}\quad \displaystyle\frac{4 \sqrt{35}}{3 \sqrt{7}}\quad & \text{(b)}\quad \displaystyle\frac{20}{\sqrt{5}} \\\\ \text{(c)}\quad \displaystyle\frac{18}{\sqrt[3]{2}}\quad& \text{(d)}\quad \displaystyle\frac{\sqrt[3]{32}}{\sqrt[4]{27}} \\\\ \text{(e)}\quad \displaystyle\frac{\sqrt[3]{36 a^{^{2}}}}{\sqrt[3]{9 a}}\quad & \text{(f)}\quad \displaystyle\frac{\sqrt[3]{2}}{\sqrt[6]{12}} \\\\ \text{(g)}\quad \displaystyle\frac{1}{\sqrt[3]{x y^{^{2}}}}\quad & \text{(h)}\quad \sqrt[m]{{\displaystyle\frac{{2{{x}^{2}}{{y}^{{3m}}}}}{{9{{x}^{5}}{{y}^{{4m-1}}}}}}} \end{array}$

Show/Hide Solution

$\begin{array}{l}\text{(a)}\ \ \ \ \displaystyle \frac{{4\sqrt{{35}}}}{{3\sqrt{7}}}\\\\\ \ \ \ \ =\displaystyle \frac{4}{3}\sqrt{{\displaystyle \frac{{35}}{7}}}\\\\\ \ \ \ \ =\displaystyle \frac{4}{3}\sqrt{{\displaystyle \frac{{7\times 5}}{7}}}\\\\\ \ \ \ \ =\displaystyle \frac{{4\sqrt{5}}}{3}\\\\\\\text{(b)}\ \ \ \ \displaystyle \frac{{20}}{{\sqrt{5}}}\\\\\ \ \ \ \ =\displaystyle \frac{{20}}{{\sqrt{5}}}\times \displaystyle \frac{{\sqrt{5}}}{{\sqrt{5}}}\ \\\\\ \ \ \ \ =\displaystyle \frac{{20\sqrt{5}}}{5}\\\\\ \ \ \ \ =4\sqrt{5}\\\\\\\text{(c)}\ \ \ \ \displaystyle \frac{{18}}{{\sqrt[3]{2}}}\\\\\ \ \ \ \ =\displaystyle \frac{{18}}{{\sqrt[3]{2}}}\times \displaystyle \frac{{\sqrt[3]{{{{2}^{2}}}}}}{{\sqrt[3]{{{{2}^{2}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{18\sqrt[3]{4}}}{{\sqrt[3]{{{{2}^{3}}}}}}\ \\\\\ \ \ \ \ =\displaystyle \frac{{18\sqrt[3]{4}}}{2}\\\ \ \ \ \ =9\sqrt[3]{4}\\\\\text{(d)}\ \ \ \ \displaystyle \frac{{\sqrt[3]{{32}}}}{{\sqrt[4]{{27}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{{{{2}^{3}}\cdot 4}}}}{{\sqrt[4]{{{{3}^{3}}}}}}\times \displaystyle \frac{{\sqrt[4]{3}}}{{\sqrt[4]{3}}}\\\\\ \ \ \ \ =\displaystyle \frac{{2\sqrt[3]{4}\sqrt[4]{3}}}{3}\\\\\\\text{(e)}\ \ \ \ \displaystyle \frac{{\sqrt[3]{{36{{a}^{2}}}}}}{{\sqrt[3]{{9a}}}}\\\\\ \ \ \ \ =\sqrt[3]{{\displaystyle \frac{{36{{a}^{2}}}}{{9a}}}}\\\\\ \ \ \ \ =\sqrt[3]{{4a}}\\\\\\\text{(f)}\ \ \ \ \ \displaystyle \frac{{\sqrt[3]{2}}}{{\sqrt[6]{{12}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{2}}}{{\sqrt[6]{{{{2}^{2}}}}\cdot \sqrt[6]{3}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{2}}}{{\sqrt[3]{2}\cdot \sqrt[6]{3}}}\times \displaystyle \frac{{\sqrt[6]{{{{3}^{5}}}}}}{{\sqrt[6]{{{{3}^{5}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[6]{{243}}}}{3}\\\\\\\text{(g)}\ \ \ \ \ \displaystyle \frac{1}{{\sqrt[3]{{x{{y}^{2}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{1}{{\sqrt[3]{{x{{y}^{2}}}}}}\times \displaystyle \frac{{\sqrt[3]{{{{x}^{2}}y}}}}{{\sqrt[3]{{{{x}^{2}}y}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{{{{x}^{2}}y}}}}{{\sqrt[3]{{{{x}^{3}}{{y}^{3}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{{{{x}^{2}}y}}}}{{xy}}\\\\\text{(h)}\ \ \ \ \ \sqrt[m]{{\displaystyle \frac{{2{{x}^{2}}{{y}^{{3m}}}}}{{9{{x}^{5}}{{y}^{{4m-1}}}}}}}\\\\\ \ \ \ \ =\sqrt[m]{{\displaystyle \frac{{2y}}{{{{3}^{2}}{{x}^{3}}{{y}^{m}}}}}}\\\\\ \ \ \ \ =\ \displaystyle \frac{{\sqrt[m]{{2y}}}}{{\sqrt[m]{{{{3}^{2}}{{x}^{3}}{{y}^{m}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[m]{{2y}}}}{{y\sqrt[m]{{{{3}^{2}}{{x}^{3}}}}}}\times \displaystyle \frac{{\sqrt[m]{{{{3}^{{m-2}}}{{x}^{{m-3}}}}}}}{{\sqrt[m]{{{{3}^{{m-2}}}{{x}^{{m-3}}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[m]{{2\cdot {{3}^{{m-2}}}{{x}^{{m-3}}}y}}}}{{3xy}}\end{array}$

6.         Reduce the order as far as possible.

$\begin{array}{ll} \text{(a)}\quad \sqrt[4]{25} \quad& \text{(b)}\quad \sqrt[6]{4} \\\\ \text{(c)}\quad \sqrt[6]{8} \quad& \text{(d)}\quad \sqrt[9]{8 y^{^{3}}} \\\\ \text{(e)}\quad \sqrt[6]{27^{^{3}}} \quad& \text{(f)}\quad \sqrt[8]{a^{2} b^{^{4}}} \\\\ \text{(g)}\quad \sqrt[12]{64 a^{^{2}} b^{^{6}}} & \text{(h)}\quad (72)^{^{\tfrac{3}{5}}} \\\\ \text{(i)}\quad \sqrt[3]{768} \end{array}$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\quad \sqrt[4]{25}=\sqrt[4]{5^{2}}=\sqrt{5}\\\\ \text{(b)}\quad \sqrt[6]{4}=\sqrt[6]{2^{2}}=\sqrt[3]{2}\\\\ \text{(c)}\quad \sqrt[6]{8}=\sqrt[6]{2^{3}}=\sqrt{2}\\\\ \text{(d)}\quad \sqrt[9]{8 y^{3}}=\sqrt[9]{2^{3} \cdot y^{3}}=\sqrt[3]{2 y}\\\\ \text{(e)}\quad \sqrt[6]{27^{3}}=\sqrt{27}=3 \sqrt{3}\\\\ \text{(f)}\quad \sqrt[8]{a^{2} b^{4}}=\sqrt[4]{a b^{2}}\\\\ \text{(g)}\quad \sqrt[12]{64 a^{2} b^{6}}=\sqrt[12]{8^{2} \cdot a^{2} b^{6}}=\sqrt[6]{8 a b^{3}}\\\\ \text{(h)}\quad (72)^{\frac{3}{5}}=\sqrt[5]{72^{3}}=\sqrt[5]{\left(3^{2} \cdot 2^{3}\right)^{3}}=\sqrt[5]{3^{6} \cdot 2^{9}}=6 \sqrt[5]{3 \cdot 16}=6 \sqrt[5]{48}\\\\ \text{(i)}\quad \sqrt[3]{768}=\sqrt[3]{4^{3} \cdot 12}=4 \sqrt[3]{12} \end{array}$

7.         Find the simplified forms.

$\begin{array}{ll} \text{(a)}\quad \sqrt{\displaystyle\frac{9}{50}} \quad& \text{(b)}\quad \sqrt[3]{\displaystyle\frac{-192}{49}} \\\\ \text{(c)}\quad \sqrt[4]{16}\quad& \text{(d)}\quad 2 \sqrt[3]{56} \end{array}$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\quad\sqrt{\displaystyle \frac{9}{50}}=\sqrt{\displaystyle \frac{3^{2}}{5^{2} \cdot 2}}=\displaystyle \frac{3}{5 \sqrt{2}} \times \displaystyle \frac{\sqrt{2}}{\sqrt{2}}=\displaystyle \frac{3 \sqrt{2}}{10}\\\\ \text{(b)}\quad\sqrt[3]{\displaystyle \frac{-192}{49}}=\sqrt[3]{\displaystyle \frac{(-4)^{3} \cdot 3}{7^{2}} \times \displaystyle \frac{7}{7}}=-\displaystyle \frac{4 \sqrt[3]{21}}{7}\\\\ \text{(c)}\quad\sqrt[4]{16}=\sqrt[4]{2^{4}}=2\\\\ \text{(d)}\quad 2 \sqrt[3]{56}=2 \sqrt[3]{2^{3} \cdot 7}=4 \sqrt[3]{7} \end{array}$

EXPONENTS AND RADICALS : EXERCISE (2.2) SOLUTIONS

1.        Evaluate the following.

         (a)     $(125)^{\frac{2}{3}}$
         (b)     $(81)^{-\frac{3}{2}}$
         (c)     $(-27)^{\frac{2}{3}}$
         (d)     $\left(\displaystyle\frac{16}{81}\right)^{-\frac{3}{4}}$
         (e)     $\left(\displaystyle\frac{-125}{8} \div \frac{1}{64}\right)^{\frac{1}{3}}$
         (f)     $(0.125)^{-\frac{2}{3}}$
         (g)     $\left(\displaystyle\frac{64}{27}\right)^{-\frac{2}{3}}$
         (h)     $(-4)^{-1}+(-1)^{-4}$

Show/Hide Solution

$ \displaystyle \begin{array}{l} \text{(a)}\ \ {{(125)}^{{\frac{2}{3}}}}={{({{5}^{3}})}^{{\frac{2}{3}}}}={{5}^{2}}=25\\\\ \text{(b)}\ \ {{(81)}^{{-\frac{3}{2}}}}={{({{9}^{2}})}^{{-\frac{3}{2}}}}={{9}^{{-3}}}=\displaystyle\frac{1}{{{{9}^{3}}}}=\displaystyle\frac{1}{{729}}\\\\ \text{(c)}\ \ {{(-27)}^{{\frac{2}{3}}}}={{\left( {{{{(-3)}}^{3}}} \right)}^{{\frac{2}{3}}}}={{(-3)}^{2}}=9\\\\ \text{(d)}\ \ {{\left( {\displaystyle\frac{{16}}{{81}}} \right)}^{{-\frac{3}{4}}}}={{\left( {{{{\left( {\frac{2}{3}} \right)}}^{4}}} \right)}^{{-\frac{3}{4}}}}={{\displaystyle\left( {\frac{2}{3}} \right)}^{{-3}}}={{\displaystyle\left( {\frac{3}{2}} \right)}^{3}}=\displaystyle\frac{{27}}{8}\\\\ \text{(e)}\ \ \ {{\left( {\displaystyle\frac{{-125}}{8}\div \frac{1}{{64}}} \right)}^{{\frac{1}{3}}}}\\\ \ \ ={{\left( {\displaystyle\frac{{-125}}{8}\times 64} \right)}^{{\frac{1}{3}}}}\\\ \ \ ={{\left( {-125\times 8} \right)}^{{\frac{1}{3}}}}\\\ \ \ ={{\left( {{{{\left( {-5} \right)}}^{3}}\times {{2}^{2}}} \right)}^{{\frac{1}{3}}}}\\\ \ \ =-5\times 2\\\ \ \ =-10\\ \text{(f)}\ \ {{(0.125)}^{{-\frac{2}{3}}}}={{\left( {{{{0.5}}^{3}}} \right)}^{{-\frac{2}{3}}}}={{\left( {0.5} \right)}^{{-2}}}={{\left( {\displaystyle\frac{1}{2}} \right)}^{{-2}}}=4\\\\ \text{(g)}\ \ {{\left( {\displaystyle\frac{{64}}{{27}}} \right)}^{{-\displaystyle\frac{2}{3}}}}={{\left( {\displaystyle\frac{{{{8}^{3}}}}{{{{3}^{3}}}}} \right)}^{{-\frac{2}{3}}}}={{\left( {\displaystyle \frac{8}{3}} \right)}^{{-2}}}={{\left( {\displaystyle \frac{3}{8}} \right)}^{2}}=\frac{9}{{64}}\\\\ \text{(h)}\ \ {{(-4)}^{{-1}}}+{{(-1)}^{{-4}}}=\displaystyle \frac{1}{{(-4)}}+\displaystyle \frac{1}{{{{{(-1)}}^{4}}}}=-\displaystyle \frac{1}{4}+1=\displaystyle \frac{3}{4} \end{array}$

2.        Simplify the following.

          (a)     $\sqrt[3]{4^{2}} \cdot 4^{\frac{2}{3}} \cdot\left(\displaystyle\frac{1}{4}\right)^{-\frac{2}{3}}$
          (b)     $\sqrt{\displaystyle\frac{512 \times 27^{-3} \times 81 \times 3^{8}}{3^{4}}}$
          (c)     $\left(\left(\displaystyle \frac{3}{4}\right)^{-4}\right)^{-0.5} \cdot \sqrt{\left(\displaystyle \frac{4}{3}\right)^{-1}} \div 16^{-0.5}$
          (d)     $(27)^{\frac{1}{4}}+\displaystyle \frac{24}{(8)^{-\frac{2}{3}}}+\frac{\sqrt[5]{2}}{(4)^{-\frac{2}{5}}}$
          (e)     $ \displaystyle \frac{(243)^{\frac{4}{5}}+(64)^{\frac{2}{3}}-(216)^{\frac{1}{3}}}{(225)^{\frac{1}{2}}-(16)^{\frac{3}{4}}}$

Show/Hide Solution

$\displaystyle \begin{array}{l} \text{(a)}\ \ \ \ \sqrt[3]{{{{4}^{2}}}}\cdot {{4}^{{\frac{2}{3}}}}\cdot {{\left( {\frac{1}{4}} \right)}^{{-\frac{2}{3}}}}\\ \ \ \ \ =\ {{4}^{{\frac{2}{3}}}}\cdot {{4}^{{\frac{2}{3}}}}\cdot {{4}^{{\frac{2}{3}}}}\\ \ \ \ \ =\ {{4}^{{\frac{6}{3}}}}\\ \ \ \ \ =\ {{4}^{2}}\\ \ \ \ \ =\ 16\\\\ \text{(b)}\ \ \ \ \sqrt{{\displaystyle \frac{{512\times {{{27}}^{{-3}}}\times 81\times {{3}^{8}}}}{{{{3}^{4}}}}}}\\ \ \ \ \ =\ \sqrt{{\displaystyle \frac{{{{2}^{9}}\times {{3}^{{-9}}}\times {{3}^{4}}\times {{3}^{8}}}}{{{{3}^{4}}}}}}\\ \ \ \ \ =\ \sqrt{{\displaystyle \frac{{{{2}^{9}}}}{3}}}\\ \ \ \ \ =\ 16\sqrt{{\displaystyle \frac{2}{3}}}\\ \ \ \ \ =\ \displaystyle\frac{{16\sqrt{6}}}{3}\\\\ \text{(c)}\ \ \ \ {{\left( {{{{\left( {\displaystyle \frac{3}{4}} \right)}}^{{-4}}}} \right)}^{{-0.5}}}\cdot \sqrt{{{{{\left( {\frac{4}{3}} \right)}}^{{-1}}}}}\div {{16}^{{-0.5}}}\\ \ \ \ \ =\ \ {{\left( {\displaystyle \frac{3}{4}} \right)}^{2}}\cdot \sqrt{{\frac{3}{4}}}\cdot {{16}^{{0.5}}}\\ \ \ \ \ =\ \ \displaystyle \frac{{{{3}^{2}}}}{{{{4}^{2}}}}\cdot \frac{{{{3}^{{\frac{1}{2}}}}}}{{{{4}^{{\frac{1}{2}}}}}}\cdot 4\\ \ \ \ \ =\ \ \displaystyle \frac{{9\sqrt{3}}}{8}\\\\ \text{(d)}\ \ \ \ {{(27)}^{{\frac{1}{4}}}}+\frac{{24}}{{{{{(8)}}^{{-\frac{2}{3}}}}}}+\frac{{\sqrt[5]{2}}}{{{{{(4)}}^{{-\frac{2}{5}}}}}}\\ \ \ \ \ =\ \ \ {{(27)}^{{\frac{1}{4}}}}+\frac{{{{2}^{3}}\times 3}}{{{{{({{2}^{3}})}}^{{-\frac{2}{3}}}}}}+\frac{{{{2}^{{\frac{1}{5}}}}}}{{{{{({{2}^{2}})}}^{{-\frac{2}{5}}}}}}\\ \ \ \ \ =\ \ \ {{(27)}^{{\frac{1}{4}}}}+\frac{{{{2}^{3}}\times 3}}{{{{2}^{{-2}}}}}+\frac{{{{2}^{{\frac{1}{5}}}}}}{{{{2}^{{-\frac{4}{5}}}}}}\\ \ \ \ \ =\ \ {{(27)}^{{\frac{1}{4}}}}+({{2}^{5}}\times 3)+{{2}^{{\frac{5}{5}}}}\\ \ \ \ \ =\ \ \ \sqrt[4]{{27}}+98\\\\ \text{(e)}\ \ \ \ \displaystyle \frac{{{{{(243)}}^{{\frac{4}{5}}}}+{{{(64)}}^{{\frac{2}{3}}}}-{{{(216)}}^{{\frac{1}{3}}}}}}{{{{{(225)}}^{{\frac{1}{2}}}}-{{{(16)}}^{{\frac{3}{4}}}}}}\\ \ \ \ \ =\ \ \ \displaystyle \frac{{{{{({{3}^{5}})}}^{{\frac{4}{5}}}}+{{{({{4}^{3}})}}^{{\frac{2}{3}}}}-{{{({{6}^{3}})}}^{{\frac{1}{3}}}}}}{{{{{({{{15}}^{2}})}}^{{\frac{1}{2}}}}-{{{({{2}^{4}})}}^{{\frac{3}{4}}}}}}\\ \ \ \ \ =\ \ \ \displaystyle \frac{{{{3}^{4}}+{{4}^{2}}-6}}{{15-{{2}^{3}}}}\\ \ \ \ \ =\ \ \ \displaystyle \frac{{81+16-6}}{{15-8}}\\ \ \ \ \ =\ \ \ \displaystyle \frac{{91}}{7}\\ \ \ \ \ =\ \ \ 13\end{array}$

3.        Simplify the following.

         (a)     $\displaystyle \frac{x-5 \sqrt{x}}{x-2 \sqrt{x}-15} \div\left(1+\frac{3}{\sqrt{x}}\right)^{-1}$
         (b)     $\sqrt[a]{\displaystyle \frac{\sqrt[b]{x}}{\sqrt[b]{x}}} \cdot \sqrt[b]{\displaystyle \frac{\sqrt[a]{x}}{\sqrt[a]{x}}} \cdot \sqrt[c]{\frac{\sqrt[a]{x}}{\sqrt[b]{x}}}$
         (c)     $\left[\displaystyle \frac{x^{m}-y^{m}}{x^{\frac{m}{2}}-y^{\frac{m}{2}}}-\displaystyle \frac{x^{m}-y^{m}}{x^{\frac{m}{2}}+y^{\frac{m}{2}}}\right]^{-2}$
         (d)     $\left(\displaystyle \frac{a^{\frac{3}{2}}}{b^{-\frac{1}{2}}}\right)^{4}\left(\displaystyle \frac{a^{-2}}{b^{3}}\right)$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\ \ \ \ \displaystyle \frac{{x-5\sqrt{x}}}{{x-2\sqrt{x}-15}}\div {{\left( {1+\displaystyle\frac{3}{{\sqrt{x}}}} \right)}^{{-1}}}\\ \ \ \ \ =\ \ \displaystyle \frac{{{{{\left( {\sqrt{x}} \right)}}^{2}}-5\sqrt{x}}}{{{{{\left( {\sqrt{x}} \right)}}^{2}}-2\sqrt{x}-15}}\div {{\left( {\displaystyle\frac{{\sqrt{x}+3}}{{\sqrt{x}}}} \right)}^{{-1}}}\\ \ \ \ \ =\ \ \displaystyle \frac{{\sqrt{x}\left( {\sqrt{x}-5} \right)}}{{\left( {\sqrt{x}-5} \right)\left( {\sqrt{x}+3} \right)}}\times \displaystyle \frac{{\sqrt{x}+3}}{{\sqrt{x}}}\\ \ \ \ \ =\ 1\\\\ \text{(b)}\ \ \ \ \sqrt[a]{{\displaystyle \frac{{\sqrt[b]{x}}}{{\sqrt[c]{x}}}}}\cdot \sqrt[b]{{\displaystyle \frac{{\sqrt[c]{x}}}{{\sqrt[a]{x}}}}}\cdot \sqrt[c]{{\displaystyle\frac{{\sqrt[a]{x}}}{{\sqrt[b]{x}}}}}\\ \ \ \ \ =\ \displaystyle \frac{{\sqrt[{ab}]{x}}}{{\sqrt[{ac}]{x}}}\cdot \displaystyle\frac{{\sqrt[{bc}]{x}}}{{\sqrt[{ab}]{x}}}\cdot \displaystyle\frac{{\sqrt[{ac}]{x}}}{{\sqrt[{bc}]{x}}}\\ \ \ \ \ =\ 1\\\\ \text{(c)}\ \ \ \ {{\left[ {\displaystyle\frac{{{{x}^{m}}-{{y}^{m}}}}{{{{x}^{{\frac{m}{2}}}}-{{y}^{{\frac{m}{2}}}}}}-\frac{{{{x}^{m}}-{{y}^{m}}}}{{{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}}}} \right]}^{{-2}}}\\ \ \ \ \ =\ {{\left[ {\displaystyle\frac{{{{{\left( {{{x}^{{\frac{m}{2}}}}} \right)}}^{2}}-{{{\left( {{{y}^{{\frac{m}{2}}}}} \right)}}^{2}}}}{{{{x}^{{\frac{m}{2}}}}-{{y}^{{\frac{m}{2}}}}}}-\displaystyle\frac{{{{{\left( {{{x}^{{\frac{m}{2}}}}} \right)}}^{2}}-{{{\left( {{{y}^{{\frac{m}{2}}}}} \right)}}^{2}}}}{{{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}}}} \right]}^{{-2}}}\\ \ \ \ \ =\ {{\left[ {\displaystyle\frac{{\left( {{{x}^{{\frac{m}{2}}}}-{{y}^{{\frac{m}{2}}}}} \right)\left( {{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}} \right)}}{{{{x}^{{\frac{m}{2}}}}-{{y}^{{\frac{m}{2}}}}}}-\displaystyle\frac{{\left( {{{x}^{{\frac{m}{2}}}}-{{y}^{{\frac{m}{2}}}}} \right)\left( {{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}} \right)}}{{{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}}}} \right]}^{{-2}}}\\ \ \ \ \ =\ {{\left[ {\left( {{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}} \right)-\left( {{{x}^{{\frac{m}{2}}}}-{{y}^{{\frac{m}{2}}}}} \right)} \right]}^{{-2}}}\\ \ \ \ \ =\ {{\left[ {{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}-{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}} \right]}^{{-2}}}\\ \ \ \ \ =\ {{\left[ {2{{y}^{{\frac{m}{2}}}}} \right]}^{{-2}}}\\ \ \ \ \ =\displaystyle\frac{1}{{4{{y}^{m}}}} \end{array}$

Exponents and Radicals : Exercise (2.1) Solutions

1.       Simplify by using the rules of exponents and name the rules used.

          (a) $\displaystyle \frac{{36{{a}^{4}}{{b}^{5}}}}{{100{{a}^{7}}{{b}^{3}}}}$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \displaystyle \frac{{36{{a}^{4}}{{b}^{5}}}}{{100{{a}^{7}}{{b}^{3}}}}\\ &=\displaystyle\frac{9}{{25}}\times \frac{1}{{{{a}^{{7-4}}}}}\times {{b}^{{5-3}}}\ \ \ \ \ (\text{Division Rule})\\ &=\displaystyle \frac{{9{{b}^{2}}}}{{25{{a}^{3}}}} \end{aligned}$

          (b) $\displaystyle \frac{27 a^{2} b^{5}}{\left(9 a^{2} b\right)^{2}}$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \displaystyle \frac{27 a^{2} b^{5}}{\left(9 a^{2} b\right)^{2}}\\ &=\displaystyle \frac{{27{{a}^{2}}{{b}^{5}}}}{{81{{a}^{4}}{{b}^{2}}}}\ \ \ \ \ (\text{Power of a Powar Rule})\\ &=\displaystyle \frac{1}{3}\times \frac{1}{{{{a}^{{4-2}}}}}\times {{b}^{{5-2}}}\ \ \ \ \ (\text{Division Rule})\\ &=\displaystyle \frac{{{{b}^{3}}}}{{3{{a}^{2}}}} \end{aligned}$

          (c) $\displaystyle \left(\frac{-135 a^{4} b^{5} c^{6}}{315 a^{6} b^{7} c^{8}}\right)^{2}$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \displaystyle {{\left( {\frac{{-135{{a}^{4}}{{b}^{5}}{{c}^{6}}}}{{315{{a}^{6}}{{b}^{7}}{{c}^{8}}}}} \right)}^{2}}\\ &=\displaystyle {{\left( {\frac{{-3}}{{7{{a}^{{6-4}}}{{b}^{{7-2}}}{{c}^{{8-6}}}}}} \right)}^{2}}\ \ \ (\text{Division Rule})\\ &=\displaystyle \frac{{{{{(-3)}}^{2}}}}{{{{{(7)}}^{2}}{{{({{a}^{2}})}}^{2}}({{b}^{5}}){{{({{c}^{2}})}}^{2}}}}\ \ (\text{Power of a Quotient Rule})\\ &= \displaystyle \frac{9}{{49{{a}^{4}}{{b}^{5}}{{c}^{4}}}}\ \ (\text{Power of a Power Rule}) \end{aligned}$

          (d) $\displaystyle \left(\frac{x^{4}}{y^{5}}\right)^{3}\left(\frac{y^{3}}{x^{2}}\right)^{2}$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \displaystyle {{\left( {\frac{{{{x}^{4}}}}{{{{y}^{5}}}}} \right)}^{3}}{{\left( {\frac{{{{y}^{3}}}}{{{{x}^{2}}}}} \right)}^{2}}\\ &=\displaystyle \left( {\frac{{{{x}^{{12}}}}}{{{{y}^{{15}}}}}} \right)\left( {\frac{{{{y}^{6}}}}{{{{x}^{4}}}}} \right)\ \ (\text{Power of a Quotient Rule})\\ &=\displaystyle \frac{{{{x}^{8}}}}{{{{y}^{9}}}}\ \ (\text{Division Rule})\\ \end{aligned}$

          (e) $\displaystyle \frac{2^{3^{2}}}{\left(2^{2}\right)^{3}}$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \displaystyle \frac{{{{2}^{{{{3}^{2}}}}}}}{{{{{\left( {{{2}^{2}}} \right)}}^{3}}}}\\ &=\displaystyle \frac{{{{2}^{9}}}}{{{{2}^{6}}}}\ (\text{Power of a Power Rule})\\ &={{2}^{3}}\ \ (\text{Division Rule})\\ &=8 \end{aligned}$

2.       Evaluate the followings.

          (a) $\displaystyle \frac{54^{2} \times 12^{3} \times 64^{2}\left(3^{2} \times 4^{3} \times 5^{2}\right)^{3}}{\left(3^{2} \times 15 \times 20^{3}\right)^{4}}$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \displaystyle \frac{{{{{54}}^{2}}\times {{{12}}^{3}}\times {{{64}}^{2}}{{{\left( {{{3}^{2}}\times {{4}^{3}}\times {{5}^{2}}} \right)}}^{3}}}}{{{{{\left( {{{3}^{2}}\times 15\times {{{20}}^{3}}} \right)}}^{4}}}}\\ &=\displaystyle \frac{{{{{\left( {{{3}^{3}}\times 2} \right)}}^{2}}\times {{{\left( {{{2}^{2}}\times 3} \right)}}^{3}}\times {{{\left( {{{2}^{6}}} \right)}}^{2}}\times {{{\left( {{{3}^{2}}\times {{2}^{6}}\times {{5}^{2}}} \right)}}^{3}}}}{{{{{\left( {{{3}^{2}}\times 3\times 5\times {{{\left( {{{2}^{2}}\times 5} \right)}}^{3}}} \right)}}^{4}}}}\\ &=\displaystyle \frac{{{{3}^{6}}\times {{2}^{2}}\times {{2}^{6}}\times {{3}^{3}}\times {{2}^{1}}^{2}\times {{3}^{6}}\times {{2}^{{18}}}\times {{5}^{6}}}}{{{{3}^{{12}}}\times {{5}^{4}}\times {{{\left( {{{2}^{2}}\times 5} \right)}}^{{12}}}}}\\ &=\displaystyle \frac{{{{3}^{{15}}}\times {{2}^{38}}\times {{5}^{6}}}}{{{{3}^{{12}}}\times {{5}^{4}}\times {{2}^{{24}}}\times {{5}^{{12}}}}}\\ &=\displaystyle \frac{{{{3}^{3}}\times {{2}^{{14}}}}}{{{{5}^{{10}}}}} \end{aligned}$

3.       Simplify.

          (a) $\displaystyle \left(\frac{3^{m}}{15^{n}}\right)^{3}\left(\frac{45^{n}}{255^{m}}\right)^{2}$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \displaystyle \left(\frac{3^{m}}{15^{n}}\right)^{3}\left(\frac{45^{n}}{255^{m}}\right)^{2}\\ &=\displaystyle {{\left( {\frac{{{{3}^{m}}}}{{{{3}^{n}}\times {{5}^{n}}}}} \right)}^{3}}{{\left( {\frac{{{{3}^{2}}^{n}\times {{5}^{n}}}}{{{{3}^{m}}\times {{5}^{m}}\times {{{17}}^{m}}}}} \right)}^{2}}\\ &=\displaystyle \left( {\frac{{{{3}^{3}}^{m}}}{{{{3}^{3}}^{n}\times {{5}^{3}}^{n}}}} \right)\left( {\frac{{{{3}^{4}}^{n}\times {{5}^{{2n}}}}}{{{{3}^{{2m}}}\times {{5}^{{2m}}}\times {{{17}}^{{2m}}}}}} \right)\\ &=\displaystyle \frac{{{{3}^{{3m+4}}}^{n}\times {{5}^{{2n}}}}}{{{{3}^{{2m+3n}}}\times {{5}^{{2m+3n}}}\times {{{17}}^{{2m}}}}}\\ &=\displaystyle \frac{{{{3}^{{m+n}}}}}{{{{5}^{{2m+n}}}\times {{{289}}^{m}}}} \end{aligned}$

          (b) $\displaystyle \left(\frac{20^{x}}{400^{y}}\right)^{2}\left(\frac{150^{y^{2}}}{180^{x}}\right)^{3}$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \displaystyle \left(\frac{20^{x}}{400^{y}}\right)^{2}\left(\frac{150^{y^{2}}}{180^{x}}\right)^{3}\\ &=\displaystyle {{\left( {\frac{{{{2}^{2}}^{x}\times {{5}^{x}}}}{{{{2}^{4}}^{y}\times {{5}^{{2y}}}}}} \right)}^{2}}{{\left( {\frac{{{{2}^{{{{y}^{2}}}}}\times {{3}^{{{{y}^{2}}}}}\times {{5}^{2}}^{{{{y}^{2}}}}}}{{{{2}^{2}}^{x}\times {{3}^{2}}^{x}\times {{5}^{x}}}}} \right)}^{3}}\\ &=\displaystyle {{\left( {\frac{{{{2}^{2}}^{x}\times {{5}^{x}}}}{{{{2}^{4}}^{y}\times {{5}^{{2y}}}}}} \right)}^{2}}{{\left( {\frac{{{{2}^{{{{y}^{2}}}}}\times {{3}^{{{{y}^{2}}}}}\times {{5}^{2}}^{{{{y}^{2}}}}}}{{{{2}^{2}}^{x}\times {{3}^{2}}^{x}\times {{5}^{x}}}}} \right)}^{3}}\\ &=\displaystyle \left( {\frac{{{{2}^{4}}^{x}\times {{5}^{{2x}}}}}{{{{2}^{8}}^{y}\times {{5}^{{4y}}}}}} \right)\left( {\frac{{{{2}^{{3{{y}^{2}}}}}\times {{3}^{{3{{y}^{2}}}}}\times {{5}^{6}}^{{{{y}^{2}}}}}}{{{{2}^{6}}^{x}\times {{3}^{6}}^{x}\times {{5}^{{3x}}}}}} \right)\\ &=\displaystyle {{2}^{{4x-8y}}}\cdot {{5}^{{2x-4y}}}\cdot {{2}^{{3{{y}^{2}}}}}^{{-6x}}\cdot {{3}^{{3{{y}^{2}}}}}^{{-6x}}\cdot {{5}^{{6{{y}^{2}}}}}^{{-3x}}\\ &=\displaystyle {{2}^{{3{{y}^{2}}}}}^{{-2x-8y}}\cdot {{27}^{{{{y}^{2}}}}}^{{-2x}}\cdot {{5}^{{6{{y}^{2}}}}}^{{-x-4y}} \end{aligned}$

          (c) $\displaystyle \frac{\left(x^{3}-y^{3}\right)(x+y)}{\left(x^{2}-y^{2}\right)^{3}}$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \displaystyle \frac{\left(x^{3}-y^{3}\right)(x+y)}{\left(x^{2}-y^{2}\right)^{3}}\\ &=\displaystyle \frac{{(x-y)({{x}^{2}}+xy+{{y}^{2}})(x+y)}}{{{{{\left( {(x-y)(x+y)} \right)}}^{3}}}}\\ &=\frac{{(x-y)({{x}^{2}}+xy+{{y}^{2}})(x+y)}}{{{{{(x-y)}}^{3}}{{{(x+y)}}^{3}}}}\\ &=\displaystyle \frac{{{{x}^{2}}+xy+{{y}^{2}}}}{{{{{(x-y)}}^{2}}{{{(x+y)}}^{2}}}} \end{aligned}$

          (d) $\displaystyle \frac{\left(x^{a-b} x^{b-c}\right)^{a}\left(\frac{x^{a}}{x^{c}}\right)^{c}}{\left(x^{b} x^{c}\right)^{a} \div\left(x^{a+c}\right)^{c}}$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \displaystyle \frac{\left(x^{a-b} x^{b-c}\right)^{a}\left(\frac{x^{a}}{x^{c}}\right)^{c}}{\left(x^{b} x^{c}\right)^{a} \div\left(x^{a+c}\right)^{c}}\\ &=\displaystyle \frac{{{{{\left( {{{x}^{{a-c}}}} \right)}}^{a}}{{{\left( {{{x}^{{a-c}}}} \right)}}^{c}}}}{{{{{\left( {{{x}^{{b+c}}}} \right)}}^{a}}\div {{{\left( {{{x}^{{a+c}}}} \right)}}^{c}}}}\\ &=\displaystyle \frac{{\left( {{{x}^{{{{a}^{2}}-ac}}}} \right)\left( {{{x}^{{ac-{{c}^{2}}}}}} \right)}}{{\left( {{{x}^{{ab+ac}}}} \right)\div \left( {{{x}^{{ac+c}}}^{{^{2}}}} \right)}}\\ &=\displaystyle \frac{{{{x}^{{{{a}^{2}}-{{c}^{2}}}}}}}{{{{x}^{{ab-{{c}^{2}}}}}}}\\ &=\displaystyle {{x}^{{{{a}^{2}}-ab}}} \end{aligned}$

4.       Evaluate the followings.

          (a)    $(-3)^{-2}$
          (b)    $-3^{-3}$
          (c)    $-2^{0}+5^{-1}$
          (d)    $(-2)^{-3}+2^{-2}-2^{-4}$
          (e)    $5^{0}-(-3)^{0}$
          (f)    $\displaystyle \frac{27^{-6}}{125^{-3}} \div \frac{9^{-2}}{25^{-4}}$
          (g)    $(-5)^{0}-(-5)^{-1}-(-5)^{-2}-(-5)^{-3}$
          (h)    $(-1)^{(-1)^{-1}}$
          (i)    $\displaystyle \frac{\left(180^{2}\right)^{-3}\left(6 \cdot 90^{-2}\right)^{3}}{\left(40^{-3}\right)^{2} \cdot 25^{-2}}$
          (j)    $\displaystyle \frac{\left(2^{-3}-3^{-2}\right)^{-1}}{\left(2^{-3}+3^{-2}\right)^{-1}}$

Show/Hide Solution

$ \displaystyle \begin{array}{l}\text{(a)}\ \ {{(-3)}^{{-2}}}=\displaystyle \frac{1}{{{{{(-3)}}^{2}}}}=\displaystyle \frac{1}{9}\\\\\text{(b)}\ \ -{{3}^{{-3}}}=-\displaystyle \frac{1}{{{{3}^{3}}}}=-\displaystyle \frac{1}{{27}}\\\\\text{(c)}\ \ -{{2}^{0}}+{{5}^{{-1}}}=-1+\displaystyle \frac{1}{5}=-\displaystyle \frac{4}{5}\\\\\text{(d)}\ \ -{{2}^{0}}+{{5}^{{-1}}}=-1+\displaystyle \frac{1}{5}=-\displaystyle \frac{4}{5}\\\\\text{(e)}\ \ {{5}^{0}}-{{(-3)}^{0}}=1-1=0\\\\\text{(f)}\ \ \ \ \displaystyle \frac{{{{{27}}^{{-6}}}}}{{{{{125}}^{{-3}}}}}\div \displaystyle \frac{{{{9}^{{-2}}}}}{{{{{25}}^{{-4}}}}}\\\ \ \ \ =\displaystyle \frac{{{{{({{3}^{3}})}}^{{-6}}}}}{{{{{({{5}^{3}})}}^{{-3}}}}}\div \displaystyle \frac{{{{{({{3}^{2}})}}^{{-2}}}}}{{{{{({{5}^{2}})}}^{{-4}}}}}\\\ \ \ \ =\displaystyle \frac{{{{5}^{9}}}}{{{{3}^{{18}}}}}\div \displaystyle \frac{{{{5}^{8}}}}{{{{3}^{4}}}}\\\ \ \ \ =\displaystyle \frac{{{{5}^{9}}}}{{{{3}^{{18}}}}}\times \displaystyle \frac{{{{3}^{4}}}}{{{{5}^{8}}}}\\\ \ \ \ =\displaystyle \frac{5}{{{{3}^{{14}}}}}\\\\\text{(g)}\ \ \ {{(-5)}^{0}}-{{(-5)}^{{-1}}}-{{(-5)}^{{-2}}}-{{(-5)}^{{-3}}}\\\ \ \ \ =1-\displaystyle \frac{1}{{(-5)}}-\displaystyle \frac{1}{{{{{(-5)}}^{2}}}}-\displaystyle \frac{1}{{{{{(-5)}}^{3}}}}\\\ \ \ \ =1+\displaystyle \frac{1}{5}-\displaystyle \frac{1}{{25}}+\displaystyle \frac{1}{{125}}\\\ \ \ \ =\displaystyle \frac{{125+25-5+1}}{{125}}\\\ \ \ \ =\displaystyle \frac{{146}}{{125}}\\\\\text{(h)}\ \ {{(-1)}^{{{{{(-1)}}^{{-1}}}}}}={{(-1)}^{{\displaystyle \frac{1}{{(-1)}}}}}=\ {{(-1)}^{{-1}}}=\displaystyle \frac{1}{{(-1)}}=-1\\\\\text{(i)}\ \ \ \ \displaystyle \frac{{{{{\left( {{{{180}}^{2}}} \right)}}^{{-3}}}{{{\left( {6\cdot {{{90}}^{{-2}}}} \right)}}^{3}}}}{{{{{\left( {{{{40}}^{{-3}}}} \right)}}^{2}}\cdot {{{25}}^{{-2}}}}}\\\ \ \ \ =\displaystyle \frac{{{{{\left( {{{{\left( {{{2}^{2}}\times {{3}^{2}}\times 5} \right)}}^{2}}} \right)}}^{{-3}}}{{{\left( {3\times 2\cdot {{{(2\times {{3}^{2}}\times 5)}}^{{-2}}}} \right)}}^{3}}}}{{{{{\left( {{{{({{2}^{3}}\times 5)}}^{{-3}}}} \right)}}^{2}}\cdot {{{({{5}^{2}})}}^{{-2}}}}}\\\ \ \ \ =\displaystyle \frac{{{{{\left( {{{2}^{4}}\times {{3}^{4}}\times {{5}^{2}}} \right)}}^{{-3}}}{{{\left( {3\times 2\times {{2}^{{-2}}}\times {{3}^{{-4}}}\times {{5}^{{-2}}}} \right)}}^{3}}}}{{{{{\left( {{{2}^{{-9}}}\times {{5}^{{-3}}}} \right)}}^{2}}\cdot {{5}^{{-4}}}}}\\\ \ \ \ =\displaystyle \frac{{{{2}^{{-12}}}\times {{3}^{{-12}}}\times {{5}^{{-6}}}\times {{3}^{3}}\times {{2}^{3}}\times {{2}^{{-6}}}\times {{3}^{{-12}}}\times {{5}^{{-6}}}}}{{{{2}^{{-18}}}\times {{5}^{{-6}}}\times {{5}^{{-4}}}}}\\\ \ \ \ =\displaystyle \frac{{{{2}^{{-15}}}\times {{3}^{{-21}}}\times {{5}^{{-12}}}}}{{{{2}^{{-18}}}\times {{5}^{{-10}}}}}\\\ \ \ \ =\displaystyle \frac{{{{2}^{3}}}}{{{{3}^{{21}}}\times {{5}^{2}}}}\\\\\text{(j)}\ \ \ \ \displaystyle \frac{{{{{\left( {{{2}^{{-3}}}-{{3}^{{-2}}}} \right)}}^{{-1}}}}}{{{{{\left( {{{2}^{{-3}}}+{{3}^{{-2}}}} \right)}}^{{-1}}}}}\\\ \ \ \ =\displaystyle \frac{{{{2}^{{-3}}}+{{3}^{{-2}}}}}{{{{2}^{{-3}}}-{{3}^{{-2}}}}}\\\ \ \ \ =\displaystyle \frac{{\displaystyle \frac{1}{{{{2}^{3}}}}+\displaystyle \frac{1}{{{{3}^{2}}}}}}{{\displaystyle \frac{1}{{{{2}^{3}}}}-\displaystyle \frac{1}{{{{3}^{2}}}}}}\\\ \ \ \ =\displaystyle \frac{{\displaystyle \frac{1}{8}+\displaystyle \frac{1}{9}}}{{\displaystyle \frac{1}{8}-\displaystyle \frac{1}{9}}}\\\ \ \ \ =\displaystyle \frac{{\displaystyle \frac{{17}}{{72}}}}{{\displaystyle \frac{1}{{72}}}}\\\ \ \ \ =17\end{array}$

5.       Simplify the followings.

          (a)    $\left(-3 a^{4}\right)\left(4 a^{-7}\right)$
          (b)    $\left(\displaystyle \frac{2 x^{-4}}{5 y^{2} z^{3}}\right)^{-2}$
          (c)    $\left(\displaystyle \frac{x^{2 m+n} x^{3(m-n)}}{x^{m-2 n} x^{2 m-n}}\right)^{-3}$
          (d)    $\left(\displaystyle \frac{2 x^{-3} y^{2}}{3^{-1} y^{3}}\right)^{2}\left(\displaystyle \frac{4 x^{-2} y^{3}}{3 x^{5}}\right)^{3} \div\left(\frac{81 x^{-2}}{y^{-3}}\right)^{-2}$
          (e)    $ \displaystyle \frac{2 x+y}{x^{-1}+2 y^{-1}}$
          (f)    $\left(x^{-2}-y^{-1}\right)^{-3}$
          (g)    $ \displaystyle \frac{x^{-2}-y^{-2}}{x^{-1}+y^{-1}}$
          (h)    $\displaystyle \frac{\left(x+y^{-1}\right)^{2}}{1+x^{-1} y^{-1}}$

Show/Hide Solution

$ \displaystyle \begin{array}{l}(\text{a})\ \ \left( {-3{{a}^{4}}} \right)\left( {4{{a}^{{-7}}}} \right)=-12{{a}^{{-3}}}=-\displaystyle \frac{{12}}{{{{a}^{3}}}}\\\\(\text{b})\ \ {{\left( {\displaystyle \frac{{2{{x}^{{-4}}}}}{{5{{y}^{2}}{{z}^{3}}}}} \right)}^{{-2}}}=\displaystyle \frac{{2{{x}^{{-8}}}}}{{5{{y}^{{-4}}}{{z}^{{-6}}}}}=\displaystyle \frac{{2{{y}^{4}}{{z}^{6}}}}{{5{{x}^{8}}}}\\\\(\text{c})\ \ {{\left( {\displaystyle \frac{{{{x}^{{2m+n}}}{{x}^{{3(m-n)}}}}}{{{{x}^{{m-2n}}}{{x}^{{2m-n}}}}}} \right)}^{{-3}}}\\\ \ \ ={{\left( {\displaystyle \frac{{{{x}^{{5m-2n}}}}}{{{{x}^{{3m-3n}}}}}} \right)}^{{-3}}}\\\ \ \ =\displaystyle \frac{{{{x}^{{-15m+6n}}}}}{{{{x}^{{-9m+9n}}}}}\\\ \ \ =\displaystyle \frac{1}{{{{x}^{{6m+3n}}}}}\\\\(\text{d})\ \ {{\left( {\displaystyle \frac{{2{{x}^{{-3}}}{{y}^{2}}}}{{{{3}^{{-1}}}{{y}^{3}}}}} \right)}^{2}}{{\left( {\displaystyle \frac{{4{{x}^{{-2}}}{{y}^{3}}}}{{3{{x}^{5}}}}} \right)}^{3}}\div {{\left( {\displaystyle \frac{{81{{x}^{{-2}}}}}{{{{y}^{{-3}}}}}} \right)}^{{-2}}}\\\ \ \ ={{\left( {\displaystyle \frac{6}{{{{x}^{3}}y}}} \right)}^{2}}{{\left( {\displaystyle \frac{{4{{y}^{3}}}}{{3{{x}^{7}}}}} \right)}^{3}}{{\left( {\displaystyle \frac{{81{{y}^{3}}}}{{{{x}^{2}}}}} \right)}^{2}}\\\ \ \ =\left( {\displaystyle \frac{{{{2}^{2}}\cdot {{3}^{2}}}}{{{{x}^{6}}{{y}^{2}}}}} \right)\left( {\displaystyle \frac{{{{2}^{6}}{{y}^{9}}}}{{{{3}^{3}}{{x}^{{21}}}}}} \right)\left( {\displaystyle \frac{{{{3}^{8}}{{y}^{6}}}}{{{{x}^{4}}}}} \right)\\\ \ \ =\displaystyle \frac{{{{2}^{8}}\cdot {{3}^{7}}\cdot {{y}^{{13}}}}}{{{{x}^{{31}}}}}\\\\(\text{e})\ \ \displaystyle \frac{{2x+y}}{{{{x}^{{-1}}}+2{{y}^{{-1}}}}}\\\ \ \ =\displaystyle \frac{{2x+y}}{{\displaystyle \frac{1}{x}+\displaystyle \frac{2}{y}}}\\\ \ \ =\displaystyle \frac{{2x+y}}{{\displaystyle \frac{{2x+y}}{{xy}}}}\\\ \ \ =(2x+y)\left( {\displaystyle \frac{{xy}}{{2x+y}}} \right)\\\ \ \ =xy\\\\(\text{f})\ \ {{\left( {{{x}^{{-2}}}-{{y}^{{-1}}}} \right)}^{{-3}}}\\\ \ \ ={{\left( {\displaystyle \frac{1}{{{{x}^{2}}}}-\displaystyle \frac{1}{y}} \right)}^{{-3}}}\\\ \ \ ={{\left( {\displaystyle \frac{{y-{{x}^{2}}}}{{{{x}^{2}}y}}} \right)}^{{-3}}}\\\ \ \ ={{\left( {\displaystyle \frac{{{{x}^{2}}y}}{{y-{{x}^{2}}}}} \right)}^{3}}\\\ \ \ =\displaystyle \frac{{{{x}^{6}}{{y}^{3}}}}{{{{{\left( {y-{{x}^{2}}} \right)}}^{3}}}}\\\\\text{(g})\ \ \displaystyle \frac{{{{x}^{{-2}}}-{{y}^{{-2}}}}}{{{{x}^{{-1}}}+{{y}^{{-1}}}}}\\\ \ \ =\displaystyle \frac{{\displaystyle \frac{1}{{{{x}^{2}}}}-\displaystyle \frac{1}{{{{y}^{2}}}}}}{{\displaystyle \frac{1}{x}+\displaystyle \frac{1}{y}}}\\\ \ \ =\displaystyle \frac{{\displaystyle \frac{{{{y}^{2}}-{{x}^{2}}}}{{{{x}^{2}}{{y}^{2}}}}}}{{\displaystyle \frac{{y+x}}{{xy}}}}\\\ \ \ =\displaystyle \frac{{{{y}^{2}}-{{x}^{2}}}}{{{{x}^{2}}{{y}^{2}}}}\times \displaystyle \frac{{xy}}{{y+x}}\\\ \ \ =\displaystyle \frac{{(y-x)(y+x)}}{{{{{(xy)}}^{2}}}}\times \displaystyle \frac{{xy}}{{y+x}}\\\ \ \ =\displaystyle \frac{{y-x}}{{xy}}\\\\(\text{h})\ \ \displaystyle \frac{{{{{\left( {x+{{y}^{{-1}}}} \right)}}^{2}}}}{{1+{{x}^{{-1}}}{{y}^{{-1}}}}}\\\ \ \ =\displaystyle \frac{{{{{\left( {x+\displaystyle \frac{1}{y}} \right)}}^{2}}}}{{1+\displaystyle \frac{1}{{xy}}}}\\\ \ \ =\displaystyle \frac{{\displaystyle \frac{{{{{\left( {xy+1} \right)}}^{2}}}}{{{{y}^{2}}}}}}{{\displaystyle \frac{{xy+1}}{{xy}}}}\\\ \ \ =\displaystyle \frac{{{{{\left( {xy+1} \right)}}^{2}}}}{{{{y}^{2}}}}\times \displaystyle \frac{{xy}}{{xy+1}}\\\ \ \ =\displaystyle \frac{{x(xy+1)}}{y}\end{array}$