Question

rationalise the denominator.. help me by solving​

Answer 1

☆Answer:-☆

x + √(x² - y²)

__________

y

➼ Solution : -

√(x+y) +√(x-y) √(x+y) + √(x-y)

___________ × ___________

√(x+y) -√(x-y) √(x+y) + √(x-y)

➛ ((√(x+y) +(√x-y))²

______________________

(√(x+y - √(x-y) ( √(x+y +√(x-y)

➛ 2x + 2 √(x² - y²)

_____________

2y

➛ x + √(x² - y²)

__________

y

For further information refer the attachment !

Answer 2

Answer:

$\rm \dfrac{x + \sqrt{x^2 - y^2}}{y}$

Explanation:

We have to rationalize:

$\rm \Longrightarrow \dfrac{\sqrt{x + y} + \sqrt{x - y}}{\sqrt{x + y} - \sqrt{x - y}}$

To rationalize the denominator, let's multiply both the numerator & denominator by the conjugate of $\rm \sqrt{x + y} - \sqrt{x - y}$ which is $\rm \sqrt{x + y} + \sqrt{x - y}$.

$\rm \Longrightarrow \dfrac{\sqrt{x + y} + \sqrt{x - y}}{\sqrt{x + y} - \sqrt{x - y}} \times \dfrac{\sqrt{x + y} + \sqrt{x - y}}{\sqrt{x + y} + \sqrt{x - y}}$

Identities applied:

(a + b) (a + b) = (a + b)²

(a + b) (a - b) = a² - b²

$\rm \Longrightarrow \dfrac{\Big(\sqrt{x + y} + \sqrt{x - y} \Big)^2}{\Big(\sqrt{x + y}\Big)^2 - \Big(\sqrt{x - y} \Big)^2}$

Identities applied:

(a + b)² = a² + b² + 2ab

$\rm \Longrightarrow \dfrac{\Big(\sqrt{x + y}\Big)^2 + \Big(\sqrt{x - y} \Big)^2 + 2\Big(\sqrt{x + y}\Big)\Big(\sqrt{x - y}\Big)}{x + y - \Big(x - y\Big)}$

Identities applied:

(a + b)(a - b) = a² - b²

$\rm \Longrightarrow \dfrac{x + y + x - y + 2\Big(\sqrt{x^2 - y^2}\Big)}{x + y - x + y}$

$\rm \Longrightarrow \dfrac{2x + 2\Big(\sqrt{x^2 - y^2}\Big)}{2y}$

Take 2 as a common integer from the numerator.

$\rm \Longrightarrow \dfrac{2\Big(x + \sqrt{x^2 - y^2}\Big)}{2y}$

2 gets cancelled.

$\rm \Longrightarrow \dfrac{x + \sqrt{x^2 - y^2}}{y}$

Hence rationalized.