Question

Simplify: $\frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$.

Answer 1

$\frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$

Lets take $\frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}$

Lets do rationalize the denominator,

Since, the R.F of $\sqrt{10}+\sqrt{3}$ is $\sqrt{10}-\sqrt{3}$

$\frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}$

$=\frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}\times \frac{\sqrt{10}-\sqrt{3}}{\sqrt{10}-\sqrt{3}}$

$=\frac{7\sqrt{3}(\sqrt{10}-\sqrt{3})}{(\sqrt{10}{)}^2-(\sqrt{3}{)}^2}$

$=\frac{7\sqrt{30}-21}{10-3}$

$=\frac{7(\sqrt{30}-3)}{7}$

$=\sqrt{30}-3$

$\Rightarrow \frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}=\sqrt{30}-3$---(1)

Lets take $\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}$

Lets rationalize the denominator,

Since, RF of $\sqrt{6}+\sqrt{5}$ is $\sqrt{6}-\sqrt{5}$

$\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}$

$=\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}\times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}$

$=\frac{2\sqrt{5}(\sqrt{6}-\sqrt{5})}{(\sqrt{6}{)}^2-(\sqrt{5}{)}^2}$

$=\frac{2\sqrt{30}-10}{6-5}$

$=2\sqrt{30}-10$

$\Rightarrow \frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}=2\sqrt{30}-10$ ---(2)

Lets take $\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$

Lets rationalize the denominator,

Since, RF of $\sqrt{15}+3\sqrt{2}$ is $\sqrt{15}-3\sqrt{2}$

$\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$

$=\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}\times \frac{\sqrt{15}-3\sqrt{2}}{\sqrt{15}-3\sqrt{2}}$

$=\frac{3\sqrt{30}-18}{(\sqrt{15}{)}^2-(3\sqrt{2}{)}^2}$

$=\frac{3(\sqrt{30}-6)}{15-18}$

$=\frac{3(\sqrt{30}-6)}{-3}$

$=6-\sqrt{30}$

$\Rightarrow \frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}=6-\sqrt{30}$---(3)

Now, from (1), (2) and (3),

$\frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$

$=\sqrt{30}-3-(2\sqrt{30}-10)-(6-\sqrt{30})$

$=\sqrt{30}-3-2\sqrt{30}+10-6+\sqrt{30}$

$=2\sqrt{30}-2\sqrt{30}+1$

$=1$

$\therefore \frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}=1$.