Question

please answer step by step correctly I'll mark as brainliest​

Answer 1

$\sf{{\Bigg ( \dfrac{3 {x}^{2}y }{9x {y}^{4} }} \Bigg) }^{2}$

Now, we can also write it as :

$\sf{\longrightarrow\dfrac{ {(3 {x}^{2}y)}^{2} }{{(9x {y}^{4}) }^{2} }}$

Now, we have find the square of the expression given in the bracket. By using distributive property :

$\sf{\longrightarrow\dfrac{ {(3)}^{2} \times {( {x}^{2} )}^{2} \times {(y)}^{2} }{ {(9)}^{2} \times {(x)}^{2} \times {( {y}^{4} )}^{2} }}$

Finding squares of the following :

$\sf{\longrightarrow\dfrac{ 9\times {( x )}^{2 \times 2} \times {y}^{2} }{81 \times {x}^{2} \times {(y)}^{4 \times 2} }}$

$\sf { \because {({a}^{m})}^{n} = {(a)}^{m \times n } }$

$\sf{\longrightarrow\dfrac{ 1 }{9} \times {x}^{4 - 2} \times {y}^{2 - 8} }$

$\sf { \because \dfrac{{a}^{m}}{{a}^{n}}= {(a)}^{m-n } }$

$\sf{\longrightarrow\dfrac{ {x}^{2} \times {y}^{ - 6} }{9} }$

$\sf{\longrightarrow\dfrac{ {x}^{2} \times 1 }{9 \times {y}^{6} } }$

$\sf { \because {a}^{-m} = \dfrac{1}{{a}^{m}} }$

$\longrightarrow\boxed{\sf \red{\dfrac{ {x}^{2} }{9 {y}^{6} } } }$

Therefore, value of $\sf{{\Bigg ( \dfrac{3 {x}^{2}y }{9x {y}^{4} }} \Bigg) }^{2}$ is $\sf { \dfrac { {x}^{2} }{ 9 {y}^{6} } }$.

More :

$\boxed{\begin{array}{cc}\bf{\dag}\:\:\underline{\text{Law of Exponents :}}\\\\\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\sf{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}\end{array}}$