Question

find a and b
1.√3−√2÷√3+√2 =a+b√2,
2.3+√7÷3−√7=a+b√7, 3.√5+√3÷√5−√3=a+b√15. ​

Answer 1

$1. \: \: \: \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } = a + b \sqrt{2} \\ \frac{\sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } = a + b \sqrt{2} \\ \frac{( { \sqrt{3} - \sqrt{2} })^{2} }{3 - 2} = a + b \sqrt{2} \\ 3 + 2 - 2 \sqrt{3} \sqrt{2} = a + b \sqrt{2} \\ 5 - 2 \sqrt{3} \sqrt{2} = a + b \sqrt{2} \\ comparing \: both \: sides \: we \: get \\ a = 5 \: \: \: and \: \: \: b = - 2 \sqrt{3} \\ \\ 2. \: \: \: \frac{3 + \sqrt{7} }{3 - \sqrt{7} } = a + b \sqrt{7} \\ \frac{( {3 + \sqrt{7} })^{2} }{ {3}^{2} - ( { \sqrt{7} })^{2} } = a + b \sqrt{7} \\ \frac{9 + 7 + 6 \sqrt{7} }{9 - 7} = a + b \sqrt{7} \\ \frac{16 + 6 \sqrt{7} }{2} = a + b \sqrt{7} \\ 8 + 3 \sqrt{7} = a + b \sqrt{7} \\ so \: \: \: a = 8 \: \: and \: \: b = 3 \\ \\ 3. \: \: \: \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } = a + b \sqrt{15} \\ \frac{( { \sqrt{5} + \sqrt{3} })^{2} }{( { \sqrt{5} })^{2} - ( { \sqrt{3} })^{2} } = a + b \sqrt{15} \\ \frac{5 + 3 + 2 \sqrt{15} }{5 - 3} = a + b \sqrt{15} \\ \frac{8 + 2 \sqrt{15} }{2} = a + b \sqrt{15} \\ 4 + \sqrt{15} = a + b \sqrt{15 } \\ comparing \: both \: sides \\ a = 4 \: \: \: b = 1$