Question

3 root 6 - 7 root 2 / 5 root 6 + 2 root 2 rationalise the denominator

Answer 1

Question

$\rm \: rationalise \: the \: denomintor \: of \: \ = \dfrac{3 \sqrt{6} - 7 \sqrt{2} }{5 \sqrt{6} + 2 \sqrt{2} }$

Solution:-

Take

$: \implies \dfrac{3 \sqrt{6} - 7 \sqrt{2} }{5 \sqrt{6} + 2 \sqrt{2} }$

$: \implies \dfrac{3 \sqrt{6} - 7 \sqrt{2} }{5 \sqrt{6} + 2 \sqrt{2} } \times \dfrac{5 \sqrt{6} - 2 \sqrt{2} }{5 \sqrt{6} - 2 \sqrt{2} }$

use this identity

:- ( a - b ) × ( a + b ) = ( a² - b² )

:- ( a - b ) × ( c - d ) = (ac - ad - bc + bd )

Applying this identity

$: \implies \dfrac{(3 \sqrt{6} - 7 \sqrt{2} )(5 \sqrt{6} - 2 \sqrt{2}) }{(5 \sqrt{6} + 2 \sqrt{2} )(5 \sqrt{6} - 2 \sqrt{2}) }$

$\rm \: : \implies \dfrac{3 \sqrt{6} \times 5 \sqrt{6} - 3 \sqrt{6} \times 2 \sqrt{2} - 7 \sqrt{2} \times 5 \sqrt{6} + 7 \sqrt{2} \times 2 \sqrt{2} }{(5 \sqrt{6}) {}^{2} - (2 \sqrt{2} ) {}^{2} }$

$\rm \: : \implies \dfrac{15 \times 6 - 6 \sqrt{12} - 35 \sqrt{12} +14 \times 2 }{25 \times 6 - 4 \times 2}$

$\rm \: : \implies \dfrac{90 - 41 \sqrt{12} + 28}{142}$

$\rm \: : \implies \dfrac{118 - 41 \sqrt{12} }{142}$

Answer:-

$\rm \: : \implies \boxed{ \dfrac{118 - 41 \sqrt{12} }{142} }$

Answer 2

Answer:

hope it will help you...