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

$\frac{1}{\sqrt{11 - 2\sqrt{30}}} - \frac{3}{\sqrt{7 - 2\sqrt{10}}} - \frac{4}{\sqrt{8 + 4\sqrt{3}}} = $

If $2^x = 4^y = 8^z$ and $xyz = 288$,then $\frac{1}{2x} + \frac{1}{4y} + \frac{1}{8z} = $

$\sqrt{2+\sqrt{5}-\sqrt{6-3 \sqrt{5}+\sqrt{14-6 \sqrt{5}}}}$ is equal to

$({x^5})^{1/3} (16{x^3})^{2/3} (\frac{1}{4} x^{4/9})^{-3/2} = $

If $a = \sqrt{21} - \sqrt{20}$ and $b = \sqrt{18} - \sqrt{17}$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module