Question

Solve $\frac{1}{\sqrt{19-\sqrt{360}}}-\frac{1}{\sqrt{21-\sqrt{440}}}+\frac{2}{\sqrt{20+\sqrt{396}}}=?$

Answer 1

Consider the first term

$\Rightarrow \frac{1}{\sqrt{19-\sqrt{360}}}$

On rationalizing, we get

$\Rightarrow \frac{\sqrt{19+\sqrt{360}}}{\sqrt{{19}^2-360}}=\sqrt{19+\sqrt{360}}$

$\Rightarrow \sqrt{19+\sqrt{360}}=\sqrt{19+2\sqrt{90}}=\sqrt{9+10+2\sqrt{9}\sqrt{10}}=\sqrt{(\sqrt{9}+\sqrt{1}0{)}^2}=\sqrt{9}+\sqrt{10}$

Now consider the second term

$\Rightarrow \frac{1}{\sqrt{21-\sqrt{440}}}$

On rationalizing, we get

$\Rightarrow \frac{\sqrt{21+\sqrt{440}}}{\sqrt{{21}^2-440}}=\sqrt{21+\sqrt{440}}$

$\Rightarrow \sqrt{21+\sqrt{440}}=\sqrt{21+2\sqrt{110}}=\sqrt{10+11+2\sqrt{10}\sqrt{11}}=\sqrt{(\sqrt{10}+\sqrt{11}{)}^2}=\sqrt{10}+\sqrt{11}$

Now consider the third term

$\Rightarrow \frac{2}{\sqrt{20-\sqrt{396}}}$

On rationalizing, we get

$\Rightarrow \frac{2\sqrt{20-\sqrt{396}}}{\sqrt{{20}^2-396}}=\sqrt{20-\sqrt{396}}$

$\Rightarrow \sqrt{20-\sqrt{396}}=\sqrt{20-2\sqrt{99}}=\sqrt{9+11-2\sqrt{9}\sqrt{11}}=\sqrt{(\sqrt{11}-\sqrt{9}{)}^2}=\sqrt{11}-\sqrt{9}$

Substituting these values in the given expression we get,

$\Rightarrow (\sqrt{9}+\sqrt{10})-(\sqrt{10}+\sqrt{11})+(\sqrt{11}-\sqrt{9})$

$\Rightarrow 0$