Question

rationalise denominator and simplify
b^2/(√,a^2+b^2)+a​

Answer 1

Concept Used :-

Method of Rationalization :-

This method of Rationalization is used to remove the radicals from denominator and it means multiply and divide by conjugate of denominator.

Let's solve the problem now!!!

$\large\underline{\bf{Solution-}}$

Given that,

$\rm :\longmapsto\:\dfrac{ {b}^{2} }{ \sqrt{ {a}^{2} + {b}^{2}} + a }$

$\rm = \: \: \:\dfrac{ {b}^{2} }{ \sqrt{ {a}^{2} + {b}^{2}} + a } \times \dfrac{ \sqrt{ {a}^{2} + {b}^{2}} - a}{\sqrt{ {a}^{2} + {b}^{2}} - a}$

$\rm = \: \: \dfrac{ {b}^{2}(\sqrt{ {a}^{2} + {b}^{2}} - a)}{ {( \sqrt{ {a}^{2} + {b}^{2}})}^{2} - {a}^{2} }$

$\: \: \: \: \: \: \: \: \: \: \red{\boxed{ \bf \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2}}}$

$\rm = \: \: \dfrac{ {b}^{2}(\sqrt{ {a}^{2} + {b}^{2}} - a)}{ \cancel {a}^{2} + {b}^{2} - \cancel {a}^{2} }$

$\rm = \: \: \dfrac{ \cancel {b}^{2}(\sqrt{ {a}^{2} + {b}^{2}} - a)}{ \cancel {b}^{2} }$

$\rm = \: \: \sqrt{ {a}^{2} + {b}^{2}} - a$

Additional Information :-

More Identities to know:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)

Answer 2

Answer:

Concept Used :-

Method of Rationalization :-

This method of Rationalization is used to remove the radicals from denominator and it means multiply and divide by conjugate of denominator.

Let's solve the problem now!!!

$\large\underline{\bf{Solution-}}$

Given that,

$⟼a2+b2+ab2</p><p></p><p>\rm = \: \: \:\dfrac{ {b}^{2} }{ \sqrt{ {a}^{2} + {b}^{2}} + a } \times \dfrac{ \sqrt{ {a}^{2} + {b}^{2}} - a}{\sqrt{ {a}^{2} + {b}^{2}} - a}=a2+b2+ab2×a2+b2−aa2+b2−a</p><p></p><p>\rm = \: \: \dfrac{ {b}^{2}(\sqrt{ {a}^{2} + {b}^{2}} - a)}{ {( \sqrt{ {a}^{2} + {b}^{2}})}^{2} - {a}^{2} }=(a2+b2)2−a2b2(a2+b2−a)</p><p></p><p>$