Question

Rationalise the denominator:
10/
2√3−√2

Answer 1

Answer:

$\Large \frac{20 \sqrt{3} + 10 \sqrt{2} }{10}$

Step-by-step explanation:

Given:-

$\Large \frac{10}{2 \sqrt{3} - \sqrt{2} }$

Solution:-

$\Large \frac{10}{2 \sqrt{3} - \sqrt{2} } \times \frac{2 \sqrt{3} + \sqrt{2} }{2 \sqrt{3} + \sqrt{2} }$

$\Large \frac{10(2 \sqrt{3} + \sqrt{2} )}{ {(2 \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} }$

$\Large \frac{20 \sqrt{3} + 10 \sqrt{2} }{12 - 2}$

$\Large \frac{20 \sqrt{3} + 10 \sqrt{2} }{10}$

hope it will helps you friend

Answer 2

Given:-

$\frac{10}{ 2\sqrt{3 - \: \sqrt{2} } } \times \frac{ 2\sqrt{3} \: + \sqrt{2} }{2 \sqrt{3} + \sqrt{2} }$

To Assume:-

$Rationalisation \: \: of \: \: the \: \: denominator \\ \frac{10}{2 \sqrt{3} - \sqrt{2} }$

Solution:-

$\frac{10}{ 2\sqrt{3 - \: \sqrt{2} } } \times \frac{ 2\sqrt{3} \: + \sqrt{2} }{2 \sqrt{3} + \sqrt{2} } \\ \frac{10(2 \sqrt{3} + \sqrt{2}) }{(2 \sqrt {3}^{2}) - ( \sqrt{2} {}^{2} } \\ \frac{20 \sqrt{3} + 10 \sqrt{2} }{12 - 2} \\ \\ \frac{20 \sqrt{3} + 10 \sqrt{2} }{10}$

So, the answer will be$\frac{20 \sqrt{3} + 10 \sqrt{2} }{10}$

Hope it helps u