Question

√2 +√3\3√2-√3=a-v√6

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Answer 1

Hey friend,
Here is the answer you were looking for:
$\frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } = a - b \sqrt{6} \\ \\ on \: rationalizing \: we \: get \\ \\ \frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } \times \frac{3 \sqrt{2} + 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } \\ \\ by \: using \: the \: identity \\ (a + b)(a - b) \\ (putting \: 3 \sqrt{2} = a \: and \: 2 \sqrt{3} = b) \\ \\ \frac{ \sqrt{2} \times 3 \sqrt{2} + \sqrt{2} \times 2 \sqrt{3} + \sqrt{3} \times 3 \sqrt{2} + \sqrt{3} \times 2 \sqrt{3} }{ {(3 \sqrt{2}) }^{2} - {(2 \sqrt{3}) }^{2} } \\ \\ = \frac{6 + 2 \sqrt{6} + 3 \sqrt{6} + 6 }{18 - 12} \\ \\ = \frac{12 + 5 \sqrt{6} }{6} \\ \\ (putting \: \frac{18 + 5 \sqrt{6} }{6} = a - b \sqrt{6} \: we \: get) \\ \\ a = \frac{12}{6} \\ \\ a = 2\\ \\ b = -\frac{5}{6}$

Hope this helps!!!!

@Mahak24

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Answer 2

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

$\frac{( \sqrt{2} + \sqrt{3}) * (3 \sqrt{2} + 2 \sqrt{3} ) }{(3 \sqrt{2} - 2 \sqrt{3}) * (3 \sqrt{2} + 2 \sqrt{3})} = a - b \sqrt{6}$

$\frac{6 + 3 \sqrt{6} + 2 \sqrt{6} + 6 }{18 - 12} = a - b \sqrt{6}$

$\frac{12 + 5 \sqrt{6} }{6} = a - b \sqrt{6}$

$2 + \frac{5 \sqrt{6} }{6} = a - b \sqrt{6}$


Here,

$a = 2, b = \frac{-5}{6}$


Hope this helps!