The expression (2+sqrt{2})+frac{1}{(2+sqrt{2} | vedclass

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

Question

(A) Given expression: $(2+\sqrt{2}) + \frac{1}{2+\sqrt{2}} + \frac{1}{2-\sqrt{2}}$First,simplify the sum of the fractions:$\frac{1}{2+\sqrt{2}} + \fra

Vedclass · Accepted Answer

$4+\sqrt{2}$

Vedclass · Answer

(A) Given expression: $(2+\sqrt{2}) + \frac{1}{2+\sqrt{2}} + \frac{1}{2-\sqrt{2}}$First,simplify the sum of the fractions:$\frac{1}{2+\sqrt{2}} + \frac{1}{2-\sqrt{2}} = \frac{(2-\sqrt{2}) + (2+\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}$Using the identity $(a+b)(a-b) = a^2 - b^2$ in the denominator:$= \frac{4}{2^2 - (\sqrt{2})^2} = \frac{4}{4-2} = \frac{4}{2} = 2$Now,add this result to the first term:$(2+\sqrt{2}) + 2 = 4+\sqrt{2}$

The expression $(2+\sqrt{2})+\frac{1}{(2+\sqrt{2})}+\frac{1}{(2-\sqrt{2})}$ equals

Similar Questions

$28 \sqrt{?} + 1426 = \frac{3}{4}$ of $2872$. The value in place of $(?)$ is

If $\sqrt{x} + \sqrt{441} = 0.02$,then the value of $x$ is

The smallest number which is a perfect square and contains $7936$ as a factor is

Find the square root of $104976$.

If $\sqrt{x}=\sqrt{3}-\sqrt{5},$ then the value of $x^{2}-16x+6$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module