Question

Find the value {x/y}^(√79+√77) ]^(√79-√77)

Answer 1

{x/y}^(√79+√77) ]^(√79-√77)  = (x/y)²

Step-by-step explanation:

{x/y}^(√79+√77) ]^(√79-√77)

as we know that

(a^p)^q  =  a^pq

=> {x/y}^(√79+√77) ]^(√79-√77)

=  {x/y}^(√79+√77) (√79-√77) )

= (x/y)⁽⁷⁹⁻⁷⁷⁾

= (x/y)²

{x/y}^(√79+√77) ]^(√79-√77)  = (x/y)²

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Answer 2

$\left(\left(\dfrac{x}{y}\right)^{\left(\sqrt{79}+\sqrt{77}\right)}\right)^{\left(\sqrt{79}-\sqrt{77}\right)}=\dfrac{x^2}{y^2}$

Step-by-step explanation:

Given: $\left(\left(\dfrac{x}{y}\right)^{\left(\sqrt{79}+\sqrt{77}\right)}\right)^{\left(\sqrt{79}-\sqrt{77}\right)}$

Using exponent law: -

$(a^m)^n=a^{mn}$

$\Rightarrow \left(\dfrac{x}{y}\right)^{(\sqrt{79}+\sqrt{77})(\sqrt{79}-\sqrt{77})}$

Using difference of square formula:-

$(a+b)(a-b)=a^2-b^2$

$\Rightarrow \left(\dfrac{x}{y}\right)^{(\sqrt{79})^2-(\sqrt{77})^2}$

cancel square root with square

$\Rightarrow \left(\dfrac{x}{y}\right)^{79-77}$

$\Rightarrow \left(\dfrac{x}{y}\right)^{2}$

$\Rightarrow \dfrac{x^2}{y^2}$

Hence, the simplest form: $\Rightarrow \dfrac{x^2}{y^2}$

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