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

  • A
    $0$
  • B
    $1$
  • C
    $\sqrt{2}$
  • D
    $1/\sqrt{2}$

Explore More

Similar Questions

જો $\frac{{({2^{n + 1}})^m}({2^{2n}}){2^n}}{{({2^{m + 1}})^n}{2^{2m}}} = 1$ હોય,તો $m =$

Difficult
View Solution

${\frac{{[4 + \sqrt{15}]}^{3/2} + {[4 - \sqrt{15}]}^{3/2}}{{[6 + \sqrt{35}]}^{3/2} - {[6 - \sqrt{35}]}^{3/2}}} = $

Difficult
View Solution

$x \ne 0$ માટે,$\left( \frac{x^l}{x^m} \right)^{(l^2 + lm + m^2)} \left( \frac{x^m}{x^n} \right)^{(m^2 + nm + n^2)} \left( \frac{x^n}{x^l} \right)^{(n^2 + nl + l^2)} = $

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

Difficult
View Solution

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

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo