Question
Prove that:
$\frac{1}{3+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+1}=1$
$\frac{1}{3+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+1}=1$
✓
Answer
$\text{L.H.S}=\frac{1}{3+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+1}$
$=\frac{1}{3+\sqrt{7}}\times\frac{3-\sqrt{7}}{3-\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}}\times\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}\\+\frac{1}{\sqrt{5}+\sqrt{3}}\times\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{1}{\sqrt{3}+1}\times\frac{\sqrt{3}-1}{\sqrt{3}+1}$
$=\frac{3-\sqrt{7}}{3^2-\big(\sqrt{7}\big)^2}+\frac{\sqrt{7}-\sqrt{5}}{\big(\sqrt{7}\big)^2-\big(\sqrt{5}\big)^2}+\frac{\sqrt{5}-\sqrt{3}}{\big(\sqrt{5}\big)^2-\big(\sqrt{3}\big)^2}+\frac{\sqrt{3}-1}{\big(\sqrt{3}\big)^2-1}$
$=\frac{3-\sqrt{7}}{9-7}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{3}-1}{3-1}$
$=\frac{3-\sqrt{7}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{3}-1}{2}$
$=\frac{3-\sqrt{7}+\sqrt{7}-\sqrt{5}+\sqrt{5}-\sqrt{3}+\sqrt{3}-1}{2}$
$=\frac{2}{2}$
$=1$
$=\text{R.H.S.}$
$=\frac{1}{3+\sqrt{7}}\times\frac{3-\sqrt{7}}{3-\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}}\times\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}\\+\frac{1}{\sqrt{5}+\sqrt{3}}\times\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{1}{\sqrt{3}+1}\times\frac{\sqrt{3}-1}{\sqrt{3}+1}$
$=\frac{3-\sqrt{7}}{3^2-\big(\sqrt{7}\big)^2}+\frac{\sqrt{7}-\sqrt{5}}{\big(\sqrt{7}\big)^2-\big(\sqrt{5}\big)^2}+\frac{\sqrt{5}-\sqrt{3}}{\big(\sqrt{5}\big)^2-\big(\sqrt{3}\big)^2}+\frac{\sqrt{3}-1}{\big(\sqrt{3}\big)^2-1}$
$=\frac{3-\sqrt{7}}{9-7}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{3}-1}{3-1}$
$=\frac{3-\sqrt{7}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{3}-1}{2}$
$=\frac{3-\sqrt{7}+\sqrt{7}-\sqrt{5}+\sqrt{5}-\sqrt{3}+\sqrt{3}-1}{2}$
$=\frac{2}{2}$
$=1$
$=\text{R.H.S.}$
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