Question

$\frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}$ find the value $a+b\sqrt{5}$

Answer 1

$\frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}=a+\sqrt{5}b$

$\frac{7+3\sqrt{5}}{3+\sqrt{5}}\times \frac{3-\sqrt{5}}{3-\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}\times \frac{3+\sqrt{5}}{3+\sqrt{5}}$

$\frac{7\left(3\sqrt{5}\right)+3\sqrt{5}(3-\sqrt{5})}{9-5}-\frac{7(3+\sqrt{5})-3\sqrt{5}(3+\sqrt{5})}{9-5}=a+\sqrt{5}b$

$\frac{21-7\sqrt{5}+9\sqrt{5}-15}{4}-\frac{21+7\sqrt{5}-9\sqrt{5}-15}{4}$ $=a+\sqrt{5}b$

$\frac{6+2\sqrt{5}}{4}-\frac{6-2\sqrt{5}}{4}=a+\sqrt{5}b$

$\frac{6+2\sqrt{5}-6+2\sqrt{5}}{4}=a+\sqrt{5}b$

$\frac{4\sqrt{5}}{4}=a+\sqrt{5}b$

$0+\sqrt{5}=a+\sqrt{5}b$

$a=0;b=1$