Question

rationalise the denominator of 1/root3-root2

Answer 1

Here's the answer.
$\sqrt{3 } + \sqrt{2}$

Answer 2

Answer:

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

Step-by-step explanation:

Consider,

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

We multiply and divide by √3 + √2

we get,

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

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

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

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

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

$\implies\sqrt{3}+\sqrt{2}$

$\implies\sqrt{3}+\sqrt{2}$