Question

Prove that 1/√2+√3-√5 +1/√2-√3-√5 = 1/√2

Answer 1

u should have to mention the brackets while asking questions

Answer 2

Answer:

We have to prove,

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

Consider,

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

$LHS=\frac{1}{(\sqrt{2}+\sqrt{3})-\sqrt{5}}+\frac{1}{(\sqrt{2}-\sqrt{3})-\sqrt{5}}$

$LHS=\frac{1}{(\sqrt{2}+\sqrt{3})-\sqrt{5}}\times\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{(\sqrt{2}+\sqrt{3})+\sqrt{5}}+\frac{1}{(\sqrt{2}-\sqrt{3})-\sqrt{5}}\times\frac{(\sqrt{2}-\sqrt{3})+\sqrt{5}}{(\sqrt{2}-\sqrt{3})+\sqrt{5}}$

$LHS=\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}+\frac{\sqrt{2}-\sqrt{3}+\sqrt{5}}{(\sqrt{2}-\sqrt{3})^2-(\sqrt{5})^2}$

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

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

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

$LHS=\frac{2\sqrt{3}+3\sqrt{2}+\sqrt{30}}{2\times6}-\frac{2\sqrt{3}-3\sqrt{2}+\sqrt{30}}{2\times6}$

$LHS=\frac{2\sqrt{3}+3\sqrt{2}+\sqrt{30}-2\sqrt{3}+3\sqrt{2}-\sqrt{30}}{12}$

$LHS=\frac{6\sqrt{2}}{12}$

$LHS=\frac{1}{\sqrt{2}}$

$LHS=RHS$

Hence Proved