Question

If a and b are rational numbers, find a and b,

* √2+√3/3√2-2√3= a+b√6
* 5+2√3/7+4√3= a+b√3

Answer 1

Answer is attached below

Hope it helps

Answer 2

Hey friend,
Here is the answer you were looking for:
$1) \\ \frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } = a + b \sqrt{6} \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } \times \frac{3 \sqrt{2} + 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } \\ \\ using \: the \: identity \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ \sqrt{2} \times 3 \sqrt{2} + \sqrt{2} \times 2 \sqrt{3} + \sqrt{3} \times 3 \sqrt{2} + \sqrt{3} \times 2 \sqrt{3} }{ {(3 \sqrt{2}) }^{2} - {(2 \sqrt{3}) }^{2} } \\ \\ = \frac{3 \times 2 + 2 \sqrt{6} + 3 \sqrt{6} + 2 \times 3}{18 - 12} \\ \\ = \frac{6 + 5 \sqrt{6} + 6}{6} \\ \\ \frac{12 + 5 \sqrt{6} }{6} = a + b \sqrt{6} \\ \\ a = \frac{12}{6} \\ \\ a = 2 \\ \\ b = \frac{5}{6} \\ \\ 2) \\ \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } = a + b \sqrt{3} \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ \\ using \: the \: identity \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{5 \times 7 - 5 \times 4 \sqrt{3} + 2 \sqrt{3} \times 7 - 2 \sqrt{3} \times 4 \sqrt{3} }{ {(7)}^{2} - {(4 \sqrt{3} )}^{2} } \\ \\ = \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 8 \times 3}{49 - 48} \\ \\ = 35 - 24 - 20 \sqrt{3} + 14 \sqrt{3} \\ \\ 11 - 6 \sqrt{3} = a + b \sqrt{3} \\ \\ a = 11 \\ \\ b = - 6$


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