Question

ration denominator ofthe following
√6/√3-√2
​

Answer 1

Answer:

Here's your answer in the attachment

Answer 2

Answer:

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

Step-by-step explanation:

For rationalizing the denominator, we multiply $\sqrt{3}+ \sqrt{2}$ with the numerator and the denominator.

$\frac{\sqrt{6}(\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})}$  = $\frac{\sqrt{18} +\sqrt{12} }{3-2}$[Using the identity (a-b)(a+b), we get $(\sqrt{3}+ \sqrt{2})(\sqrt{3} - \sqrt{2}) = (\sqrt{3})^2-(\sqrt{2})^2 = 3-2$ =1 ]

= $\sqrt{18} + \sqrt{12}$ = $3\sqrt{2} + 2\sqrt{3}$