Question

Please solve the 2nd question

Answer 1

$\frac{ \sqrt{2} - \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } = a - b \sqrt{6} \\ \frac{ \sqrt{2} - \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } \times \frac{3 \sqrt{2 } + 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } = a - b \sqrt{6} \\ \frac{6 + 2 \sqrt{6} - 3 \sqrt{6} - 6 }{(3 \sqrt{2}) ^{2} - (2 \sqrt{3}) ^{2} } = a - b \sqrt{6} \\ \frac{ - \sqrt{6} }{18 - 12} = a - b \sqrt{6} \\ \frac{ - \sqrt { 6} }{6} = a - b \sqrt{6} \: \\ so \: \: \: \: \: \: a = 0 \: \: \: \: and \: \: \: \: \: \: \: b = \frac{1}{6}$

Answer 2

Hey! Your Answer Is In Image...