Question

On rationalosing the denominator of 1/√3-√2 we get
a.)1/√3+√2
b.)√3+√2
c.)√2-√3
d.)-√3-√2
plzz fast​

Answer 1

$\tt \dfrac{1}{\sqrt{3} - \sqrt{2}}$

To rationalise a denominator, we multiply the numerator and the denominator by the conjugate of denominator.

Conjugate of √3 - √2 is √3 + √2

$\tt \dfrac{1}{\sqrt{3} - \sqrt{2}} \times \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3}+\sqrt{2}} = \dfrac{\sqrt{3} + \sqrt{2}}{ ( \sqrt{3}-\sqrt{2} ) ( \sqrt{3}+\sqrt{2} )}$

Since, (a+b)(a-b)=a² - b², ∴

$\tt\dashrightarrow \dfrac{\sqrt{3} + \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} = \dfrac{\sqrt{3} + \sqrt{2}}{3 - 2}$

$\tt\dashrightarrow \dfrac{\sqrt{3} + \sqrt{2}}{1}= \sqrt{3}+\sqrt{2}$

Hence, Rationalised.

Answer :- √3 + √2

Hence, option b is correct.