Question

pls solve this question

Answer 1

Answer:

Option(D)

Step-by-step explanation:

$Given:[(\frac{a}{b})^{\sqrt{99}-\sqrt{97}}]^{\sqrt{99}+\sqrt{97}}$

$= [\frac{a}{b}]^{(\sqrt{99}-\sqrt{97})(\sqrt{99}+\sqrt{97})}$

We know that (a-b)(a+b)= a² - b²

$=[(\frac{a}{b})]^{99-97}$

$=[\frac{a}{b}]^2$

$= \boxed{\frac{a^2}{b^2}}$

(or)

$Given:((\frac{a}{b})^{\sqrt{99}-\sqrt{97}})^{\sqrt{99}+\sqrt{97}}$

$=((\frac{a}{b})^{3\sqrt{11}-\sqrt{97})}^{\sqrt{99}+\sqrt{97}}$

$=((\frac{a}{b})^{3\sqrt{11}-\sqrt{97}})^{3\sqrt{11}+\sqrt{97}}$

$=((\frac{a}{b})^{(3\sqrt{11}-\sqrt{97})(3\sqrt{11}+\sqrt{97})}$

$=(\frac{a}{b})^{(3\sqrt{11})^2 - (\sqrt{97})^2}$

$=(\frac{a}{b})^{99-97}$

$=(\frac{a}{b})^{2}$

$=\frac{a^2}{b^2}$

Therefore, the answer is a²/b²

Hope it helps!