sqrt{6 + 2sqrt{3} + 2sqrt{2} + 2sqrt{6}} - fr | vedclass

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

Question

(D) First,simplify the term $\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}}$.This is of the form $\sqrt{a^2 + b^2 + c^2 + 2ab + 2bc + 2ca} = a + b + c$.

Vedclass · Accepted Answer

$	ext{None of these}$

Vedclass · Answer

(D) First,simplify the term $\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}}$.This is of the form $\sqrt{a^2 + b^2 + c^2 + 2ab + 2bc + 2ca} = a + b + c$.Here,$a^2 = 1, b^2 = 2, c^2 = 3$,so $a=1, b=\sqrt{2}, c=\sqrt{3}$.Thus,$\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}} = 1 + \sqrt{2} + \sqrt{3}$.Next,simplify $\frac{1}{\sqrt{5 + 2\sqrt{6}}}$.Note that $5 + 2\sqrt{6} = 3 + 2 + 2\sqrt{3 	imes 2} = (\sqrt{3} + \sqrt{2})^2$.So,$\frac{1}{\sqrt{5 + 2\sqrt{6}}} = \frac{1}{\sqrt{3} + \sqrt{2}}$.Rationalizing the denominator: $\frac{1}{\sqrt{3} + \sqrt{2}} 	imes \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2}$.Finally,calculate the expression: $(1 + \sqrt{2} + \sqrt{3}) - (\sqrt{3} - \sqrt{2}) = 1 + \sqrt{2} + \sqrt{3} - \sqrt{3} + \sqrt{2} = 1 + 2\sqrt{2}$.

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

Similar Questions

The value of the fifth root of $10^{10^{10}}$ is

If $\frac{{({2^{n + 1}})^m}({2^{2n}}){2^n}}{{({2^{m + 1}})^n}{2^{2m}}} = 1$,then $m =$

Let $\frac{7}{2^{1/2} + 2^{1/4} + 1} = A + B \cdot 2^{1/4} + C \cdot 2^{1/2} + D \cdot 2^{3/4}$,then find the value of $A + B + C + D$.

If $x = \sqrt{7 + 4\sqrt{3}}$,then $x + \frac{1}{x} = ......$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module