Question

(√2+√3)/(3√2-2√3)=a+b√6

Answer 1

$\frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } \times \frac{3 \sqrt{2} + 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } = a + b \sqrt{6} \\ \frac{3 \sqrt{2} \sqrt{2}+2\sqrt{3}\sqrt{2} +3 \sqrt{3}\sqrt{2}+2\sqrt{3} \sqrt{3} }{( {3 \sqrt{2} )}^{2} - ( {2 \sqrt{3} })^{2} } = a + b \sqrt{6} \\ \frac{6 + 2 \sqrt{6} + 3 \sqrt{6} + 6 }{18 - 12} = a + b \sqrt{6} \\ \frac{12 + 5 \sqrt{6} }{ - 6} = a + b \sqrt{6} \\ \frac{12}{ - 6} + \frac{5 \sqrt{6} }{ - 6} = a + b \sqrt{6} \\ - 2 - \frac{ 5}{6} = a + b \sqrt{6} \\ compare \: both \: equations \\ a = - 2 \\ b = \frac{ - 5}{6}$