Evaluate: frac{sqrt{6 + 2sqrt{3} + 2sqrt{2} + | vedclass

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

Question

(A) First,simplify the numerator term $\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}}$.Note that $(1 + \sqrt{2} + \sqrt{3})^2 = 1^2 + (\sqrt{2})^2 + (\s

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) First,simplify the numerator term $\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}}$.Note that $(1 + \sqrt{2} + \sqrt{3})^2 = 1^2 + (\sqrt{2})^2 + (\sqrt{3})^2 + 2(1)(\sqrt{2}) + 2(1)(\sqrt{3}) + 2(\sqrt{2})(\sqrt{3}) = 1 + 2 + 3 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6} = 6 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6}$.Thus,$\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}} = 1 + \sqrt{2} + \sqrt{3}$.Next,simplify the denominator $\sqrt{5 + 2\sqrt{6}}$.Note that $(\sqrt{3} + \sqrt{2})^2 = 3 + 2 + 2\sqrt{6} = 5 + 2\sqrt{6}$.Thus,$\sqrt{5 + 2\sqrt{6}} = \sqrt{3} + \sqrt{2}$.Substituting these into the expression: $\frac{(1 + \sqrt{2} + \sqrt{3}) - 1}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = 1$.

Evaluate: $\frac{\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}} - 1}{\sqrt{5 + 2\sqrt{6}}}$

Similar Questions

$\sqrt{-8 - 6i} = $

Let $z = i^{2i}$,then $|z|$ is (where $i = \sqrt{-1}$)

$\operatorname{sech}^{-1}\left(\frac{3}{5}\right)-\tanh ^{-1}\left(\frac{3}{5}\right)=$

By taking $\sqrt{a \pm i b}=x \pm i y, x>0$,if we get $\frac{\sqrt{21+12 \sqrt{2} i}}{\sqrt{21-12 \sqrt{2} i}}=a+i b$,then $\frac{b}{a}=$

If $\theta = \frac{\pi}{12}$ and $x = \log \left(\cot \left(\frac{\pi}{4} + \theta\right)\right)$,then $\cosh x =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module