Question

Find the value of
a and b
Solution please ​

Answer 1

Answer:

I am not understanding of your questions

Answer 2

Answer:

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

Step-by-step explanation:

$\frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } = a - b \sqrt{6} \\ = > \frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } \times \frac{ 3\sqrt{2} + 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } = a + b \sqrt{6} \\ = > \frac{ \sqrt{2} + \sqrt{3} (3 \sqrt{2} + 2 \sqrt{3} }{(3 \sqrt{2) ^{2} } - {(2 \sqrt{3)} }^{2} } = a + b \sqrt{6} \\ = > \frac{ \sqrt{2} (3 \sqrt{2} + 2 \sqrt{3} ) + \sqrt{3} (3 \sqrt{2} + 2 \sqrt{3} }{(9 \times 2) - (4 \times 3)} = a + b \sqrt{6} \\ = > \frac{6 + 2 \sqrt{6} + 3 \sqrt{6} + 6 }{18 - 12} = a + b\sqrt{6} \\ = > \frac{12 + 2 \sqrt{6} + 3 \sqrt{6} }{6} = a+ b \sqrt{6} \\ = > \frac{12 + \sqrt{6} (2 + 3)}{6} = a+ b\sqrt{6} \\ = > \frac{12 + \sqrt{6} \times 5}{6} = a + b \sqrt{6} \\ = > \frac{12 + 5 \sqrt{6} }{6} =a + b\sqrt{6} \\ = > \frac{12}{6} + \frac{5 \sqrt{6} }{6} = a + b \sqrt{6} \\ = > 2 + \frac{5}{6} \sqrt{6} = a + b \sqrt{6} \\ = > a = 2 \\ b = \frac{5}{6}$

hope it helps.