Question

Rationalise the denominator 1/(root3-root2-1)

Answer 1

hry mate, here is ur answer
hope it helps u...

Answer 2

Answer:

$\frac{1}{\sqrt{3}-\sqrt{2}-1}</p><p>=\frac{\sqrt{6}+2+\sqrt{2}}{-4}$

Step-by-step explanation:

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

Multiply numerator and denominator by [√3+(√2+1)] ,we get

$=\frac{\sqrt{3}+\sqrt{2}+1}{[\sqrt{3}-(\sqrt{2}+1)][\sqrt{3}+(\sqrt{2}+1)]}$

/* By algebraic identity:

$\boxed {(a-b)(a+b)=a^{2}-b^{2}}$*/

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

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

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

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

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

Multiply numerator and denominator by √2 , we get

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

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

Therefore,

$\frac{1}{\sqrt{3}-\sqrt{2}-1}</p><p>=\frac{\sqrt{6}+2+\sqrt{2}}{-4}$

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