Question

7+3√5\3+√5 - 7-3√5\3-√5= a+√5b

Answer 1

hope this helps you...

Answer 2

Answer:

$a = 0 \: and \: b = 1$

Step-by-step explanation:

$LHS=\frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}$$=\frac{(7+3\sqrt{5})(3-\sqrt{5})-(7-3\sqrt{5})(3+\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}$$=\frac{21-7\sqrt{5}+9\sqrt{5}-15-(21+7\sqrt{5}-9\sqrt{5}-15)}{3^{2}-(\sqrt{5})^{2}}$

$=\frac{21-7\sqrt{5}+9\sqrt{5}-15-21-7\sqrt{5}+9\sqrt{5}+15)}{9-5}$

$=\frac{4\sqrt{5}}{4}$

$=\sqrt{5}---(1)$

$RHS = a+\sqrt{5}b--(2)$

$Now\\0+\sqrt{5}=a+\sqrt{5}\\from\:(1)\:and\:(2)$

$a = 0 \: and \: b = 1$

Therefore,

$a = 0 \: and \: b = 1$

•••♪