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}$