Question

rationalise the denominator 3+√2 / (3√2-2√3)​

Answer 1

${\huge{\boxed{\red{A}}}{\boxed{\green{N}}}{\boxed{\purple{S}}}{\boxed{\orange{W}}}{\boxed{\red{E}}}{\boxed{\green{R}}}}$

To Solve:

• rationalise the denominator 3+√2 / (3√2-2√3)

Solⁿ:

${ \tt{\frac{3 + \sqrt{2} }{3 \sqrt{2} - 2 \sqrt{3} }}} \\ \\ { \tt{\frac{3 + \sqrt{2} }{3 \sqrt{2} - 2 \sqrt{3} }}} \: \times \: { \tt{\frac{3 \sqrt{2} + 2 \sqrt{3}}{ 3 \sqrt{2} + 2 \sqrt{3}}}} \\ \\ { \tt{identity = (x - y)(x + y) = {x}^{2} - {y}^{2} }} \\ \\{ \tt{ \frac{3 + \sqrt{2} \: (3 \sqrt{2} + 2 \sqrt{3)} }{ {(3 \sqrt{2} )}^{2} \: - \: {(2 \sqrt{3})}^{2} }}} \\ \\ { \tt{ \frac{6 + 9 \sqrt{2} + 6 \sqrt{3} + 2 \sqrt{6} }{(9 \times 2) -(4 \times 3)}}} \\ \\ { \boxed{ \red{ \tt{ \frac{6 + 9 \sqrt{2} + 6 \sqrt{3} + 2 \sqrt{6} }{6}}}}}$

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