frac{4}{{1 + sqrt 2 - sqrt 3 }} = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To rationalize the denominator,we write the expression as $\frac{4}{(1 + \sqrt{2}) - \sqrt{3}}$.Multiply the numerator and denominator by $(1 + \s

Vedclass · Accepted Answer

$2 + \sqrt 2 + \sqrt 6 $

Vedclass · Answer

(A) To rationalize the denominator,we write the expression as $\frac{4}{(1 + \sqrt{2}) - \sqrt{3}}$.Multiply the numerator and denominator by $(1 + \sqrt{2}) + \sqrt{3}$:$\frac{4((1 + \sqrt{2}) + \sqrt{3})}{((1 + \sqrt{2}) - \sqrt{3})((1 + \sqrt{2}) + \sqrt{3})}$$= \frac{4(1 + \sqrt{2} + \sqrt{3})}{(1 + \sqrt{2})^2 - (\sqrt{3})^2}$$= \frac{4(1 + \sqrt{2} + \sqrt{3})}{1 + 2 + 2\sqrt{2} - 3}$$= \frac{4(1 + \sqrt{2} + \sqrt{3})}{2\sqrt{2}}$$= \frac{2(1 + \sqrt{2} + \sqrt{3})}{\sqrt{2}}$$= \sqrt{2}(1 + \sqrt{2} + \sqrt{3})$$= \sqrt{2} + 2 + \sqrt{6}$$= 2 + \sqrt{2} + \sqrt{6}$.

$\frac{4}{{1 + \sqrt 2 - \sqrt 3 }} = $

Similar Questions

If $x = 3 - \sqrt{5}$,then find the value of $\frac{\sqrt{x}}{\sqrt{2} + \sqrt{3x - 2}}$.

$20^{2-3x^2} = (40\sqrt{5})^{3x^2-2}$,then $x$ is equal to

The number of digits in the decimal expansion of $16^5 \times 5^{16}$ is

If $x = 3 - \sqrt{5}$,then $\frac{\sqrt{x}}{\sqrt{2} + \sqrt{3x - 2}} = $

$\frac{3\sqrt{2}}{\sqrt{6} + \sqrt{3}} - \frac{4\sqrt{3}}{\sqrt{6} + \sqrt{2}} + \frac{\sqrt{6}}{\sqrt{3} + \sqrt{2}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module