Question

find the value of A plzz help​

Answer 1

Answer:

$\sf{2xa=(x+ \dfrac{x^{2}}{y^{2}})^{2}-(x-\dfrac{x^{2}}{y^{2}})^{2}}$

$\implies\sf{xa=x^{2}+\dfrac{x^{4}}{y^{4}}}$

$\therefore\sf{a=x+\dfrac{x^{3}}{y^{4}}}$

Answer 2

${\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}$

Given:

• $2xa = \left(x + \dfrac{x^2}{y^2}\right)^2 - \left(x - \dfrac{x^2}{y^2}\right)^2$

To find:

• Value of "a".

Solution:

$2xa = \left(x + \dfrac{x^2}{y^2}\right)^2 - \left(x - \dfrac{x^2}{y^2}\right)^2 \\ \\ We\:know\:that, \:\:\:\:\:p^2 - q^2 = (p+q)*(p-q)\\ \\So,\\ \\ \implies 2xa = \left(x + \dfrac{x^2}{y^2}\!\!\!\!\bold\Large{{/}} + x - \dfrac{x^2}{y^2} \!\!\!\!\bold\Large{{/}} \right) \times \left(x \!\!\!\!\bold{/} + \dfrac{x^2}{y^2} - x \!\!\!\!\bold{/} + \dfrac{x^2}{y^2}\right) \\ \\ \implies 2\!\!\!\bold{/}x\!\!\!\bold{/}a = 2\!\!\!\bold{/}x\!\!\!\bold{/} \times \left(\dfrac{2x^2}{y^2} \right) \\ \\ \implies \bold{ a = \dfrac{2x^2}{y^2} }$

Formula Used:

• $p^2 - q^2 = (p+q)*(p-q)$

Hope it helps!