Question

If a=√2+1 and b=√2-1, then the value of 1/(a+1)+ 1/(b-1) is-----​

Answer 1

Answer:

The answer is $-\sqrt 2$

Step-by-step explanation:

Given, $a=\sqrt 2+1$

$\Rightarrow \frac{1}{a+1}=\frac{1}{\sqrt 2+2}\cdot \frac{2-\sqrt 2}{2-\sqrt 2}\\=\frac{2-\sqrt 2}{2^2-(\sqrt 2)^2}=\frac{2-\sqrt 2}{4-2}=\frac{2-\sqrt 2}{2}$

And $b=\sqrt 2+1$

$\Rightarrow \frac{1}{b-1}=\frac{1}{\sqrt 2-2}=\frac{1}{\sqrt 2-2}\cdot \frac{\sqrt 2+2}{\sqrt 2+2}\\=\frac{\sqrt 2+2}{(\sqrt 2)^2-2^2}=\frac{\sqrt 2+2}{2-4}=\frac{\sqrt 2+2}{-2}=\frac{-2-\sqrt 2}{2}$

Thus

$\frac{1}{a+1}+\frac{1}{b-1}=\frac{2-\sqrt 2}{2}+\frac{-2-\sqrt 2}{2}=\frac{2-\sqrt 2-2-\sqrt 2}{2}=\frac{-2\sqrt 2}{2}=-\sqrt 2$