Question

Rationalise the denominator of 1/√3-√2+1
please dont write nonsense ​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\dfrac{1}{\sqrt{3}-\sqrt{2}+1}$

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

$=\dfrac{\sqrt{3}+1+\sqrt{2}}{\left(\sqrt{3}+1-\sqrt{2}\right)\left(\sqrt{3}+1+\sqrt{2}\right)}$

$=\dfrac{\sqrt{3}+1+\sqrt{2}}{\left(\sqrt{3}+1\right)^2-\left(\sqrt{2}\right)^2}$

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

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

$=\dfrac{\sqrt{3}+\sqrt{2}+1}{2+2\sqrt{3}}$

$=\dfrac{\left(\sqrt{3}+\sqrt{2}+1\right)\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}$

$=\dfrac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)+\sqrt{2}\left(\sqrt{3}-1\right)}{2\left\{\left(\sqrt{3}\right)^2-\left(1\right)^2\right\}}$

$=\dfrac{(3-1)+\sqrt{6}-\sqrt{2}}{2\left\{3-1\right\}}$

$=\dfrac{2+\sqrt{6}-\sqrt{2}}{2\cdot2}$

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