Question

simplify 1 / √3+√2-√5

Answer 1

Hey there!
Okay so basically this question is very easy. You just need to solve the question to get the radical sign out of the denominator. For the same, you just have to multiply the whole fraction by the conjugate of the number in the denominator. This process is called rationalisation. 
Let us suppose that,
a=$\sqrt{3} + \sqrt{2}$ and b= $\sqrt{5}$
We will be using the following algebraic identity,
$a^{2} - b^{2} = (a+b)(a-b)$

So, I / $\sqrt{3} + \sqrt{2}$ +  $\sqrt{5}$ x $\sqrt{3} + \sqrt{2}$ -  $\sqrt{5}$ / $\sqrt{3} + \sqrt{2}$ - $\sqrt{5}$
After solving, it becomes - 
$\sqrt{3} + \sqrt{2}$ +  $\sqrt{5}$ / 2$\sqrt{6}$

Above, I said that no radical sign should be there in the denominator. So we will again multiply it with its conjugate.
$\sqrt{3} + \sqrt{2}$ +  $\sqrt{5}$ / 2$\sqrt{6}$ x $\sqrt{6}$ / $\sqrt{6}$ 
Now the final answer is,
3$\sqrt{2} + 2 \sqrt{3} + \sqrt{30}$ / 12

Hope it helped! It took me a while to complete this answer with radical signs, but I hope it is understandable, right?

Answer 2

Answer:

$\frac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}=\frac{3\sqrt{2}+2\sqrt{3}+\sqrt{30}}{12}$

Step-by-step explanation:

Given : Expression  $\frac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}$

To find : Simplify the expression ?

Solution :

We can write expression as,

$\frac{1}{(\sqrt{3}+\sqrt{2})-\sqrt{5}}$

Rationalize the denominator,

$=\frac{1}{(\sqrt{3}+\sqrt{2})-\sqrt{5}}\times \frac{(\sqrt{3}+\sqrt{2})+\sqrt{5}}{(\sqrt{3}+\sqrt{2})+\sqrt{5}}$

$=\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{(\sqrt{3}+\sqrt{2})^2-(\sqrt{5})^2}$

$=\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{3+2\sqrt6+2-5}$

$=\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{2\sqrt6}$

Multiply and divide by $\sqrt6$

$=\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{2\sqrt6}\times \frac{\sqrt6}{\sqrt6}$

$=\frac{\sqrt{18}+\sqrt{12}+\sqrt{30}}{2\times 6}$

$=\frac{\sqrt{18}+\sqrt{12}+\sqrt{30}}{12}$

or $=\frac{3\sqrt{2}+2\sqrt{3}+\sqrt{30}}{12}$

Therefore, $\frac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}=\frac{3\sqrt{2}+2\sqrt{3}+\sqrt{30}}{12}$