frac{sqrt{2}}{sqrt{2 + sqrt{3}} - sqrt{2 - sq | vedclass

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

Question

(B) Let $x = \frac{\sqrt{2}}{\sqrt{2 + \sqrt{3}} - \sqrt{2 - \sqrt{3}}}$.Multiply the numerator and denominator by $\sqrt{2}$:$x = \frac{\sqrt{2} 	im

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Let $x = \frac{\sqrt{2}}{\sqrt{2 + \sqrt{3}} - \sqrt{2 - \sqrt{3}}}$.Multiply the numerator and denominator by $\sqrt{2}$:$x = \frac{\sqrt{2} 	imes \sqrt{2}}{\sqrt{2}(\sqrt{2 + \sqrt{3}} - \sqrt{2 - \sqrt{3}})}$$x = \frac{2}{\sqrt{4 + 2\sqrt{3}} - \sqrt{4 - 2\sqrt{3}}}$Note that $4 + 2\sqrt{3} = (\sqrt{3} + 1)^2$ and $4 - 2\sqrt{3} = (\sqrt{3} - 1)^2$.Substituting these values:$x = \frac{2}{(\sqrt{3} + 1) - (\sqrt{3} - 1)}$$x = \frac{2}{\sqrt{3} + 1 - \sqrt{3} + 1}$$x = \frac{2}{2} = 1$.

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

Similar Questions

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

The least number among $\sqrt[3]{4}, \sqrt[4]{5}, \sqrt[4]{7}$ and $\sqrt[3]{8}$ is:

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

If $3^x - 3^{x - 1} = 6$,then $x^x$ is equal to

$\frac{\sqrt{8+\sqrt{28}}+\sqrt{8-\sqrt{28}}}{\sqrt{8+\sqrt{28}}-\sqrt{8-\sqrt{28}}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module