Question

simplify under root a square minus b square + a upon under root a square + b square + b divided by under root a square + b square minus b Upon A minus under root a square + b square​

Answer 1

Answer:

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

Step-by-step explanation:

Formula used:

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

$\frac{\sqrt{a^2+b^2}+a}{\sqrt{a^2+b^2}+b}$ divided by $\frac{\sqrt{a^2+b^2}-b}{a-\sqrt{a^2+b^2}}$

$=\frac{a+\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}+b}*\frac{a-\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}-b}$

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

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

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

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

Answer 2

Given :  $\dfrac{\sqrt{A+B} +\sqrt{A-B}}{\sqrt{A+B} -\sqrt{A-B}}$

To Find : Simplify

Solution :

$\dfrac{\sqrt{A+B} +\sqrt{A-B}}{\sqrt{A+B} -\sqrt{A-B}}$

Rationalize

$\dfrac{\sqrt{A+B} +\sqrt{A-B}}{\sqrt{A+B} -\sqrt{A-B}} \times \dfrac{\sqrt{A+B} +\sqrt{A-B}}{\sqrt{A+B} +\sqrt{A-B}}$

use identity

(x + y)² = x²  + y² + 2xy

(x + y)(x - y) = x² - y²

(√x)² = x  

={ (A + B) + (A - B) + 2√(A² - B²) } / {A + B - (A - B) }

= {2A + 2√(A² - B²) } / 2B

=   { A +  √(A² - B²) } /  B

= $\dfrac{A +\sqrt{A^2-B^2}}{B}$

$\dfrac{\sqrt{A+B} +\sqrt{A-B}}{\sqrt{A+B} -\sqrt{A-B}} = \dfrac{A +\sqrt{A^2-B^2}}{B}$

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