Question

rationalize 1/root 3-root2-1

Answer 1

FIRST TAKE ROOT3-ROOT2 IN BRACKET AND -1 IN OTHER .
NOW RATIONALISE IT. NOW U GET 4-2ROOT 6 IN DENOMINATOR , RATIONALISE IT AGAIN.

Answer 2

Answer:

$\frac{1}{\sqrt{3}-\sqrt{2}-1}=\frac{2+\sqrt{2}+\sqrt{6}}{-4}$

Step-by-step explanation:

Given Expression , $\frac{1}{\sqrt{3}-\sqrt{2}-1}$

We have to rationalize it.

Consider,    

$\frac{1}{\sqrt{3}-\sqrt{2}-1}$

$\implies\frac{1}{(\sqrt{3}-1)-\sqrt{2}}$

$\implies\frac{1}{(\sqrt{3}-1)-\sqrt{2}}\times\frac{(\sqrt{3}-1)+\sqrt{2}}{(\sqrt{3}-1)+\sqrt{2}}$

$\implies\frac{\sqrt{3}-1+\sqrt{2}}{((\sqrt{3}-1)-\sqrt{2})((\sqrt{3}-1)+\sqrt{2})}$

$\implies\frac{\sqrt{3}-1+\sqrt{2}}{(\sqrt{3}-1)^2-(\sqrt{2})^2}$

$\implies\frac{\sqrt{3}-1+\sqrt{2}}{(\sqrt{3})^2+1-2\sqrt{3}-2}$

$\implies\frac{\sqrt{3}-1+\sqrt{2}}{2-2\sqrt{3}}$

$\implies\frac{\sqrt{3}-1+\sqrt{2}}{2-2\sqrt{3}}\times\frac{2+2\sqrt{3}}{2+2\sqrt{3}}$

$\implies\frac{(\sqrt{3}-1+\sqrt{2})(2+2\sqrt{3})}{(2+2\sqrt{3})(2-2\sqrt{3})}$

$\implies\frac{2\sqrt{3}+2\times3-2-2\sqrt{3}+2\sqrt{2}+2\sqrt{6}}{4-(2\sqrt{3})^2}$

$\implies\frac{4+2\sqrt{2}+2\sqrt{6}}{4-12}$

$\implies\frac{4+2\sqrt{2}+2\sqrt{6}}{-8}$

$\implies\frac{2+\sqrt{2}+\sqrt{6}}{-4}$

Therefore, $\frac{1}{\sqrt{3}-\sqrt{2}-1}=\frac{2+\sqrt{2}+\sqrt{6}}{-4}$