Question

If $x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ and $y=1$, the value of $\frac{x-y}{x-3y}$ is

• A. $-1\sqrt{3}(\sqrt{2}+\sqrt{3})$
• B. $\frac{5}{\sqrt{6}-4}$
• C. $\frac{5}{\sqrt{6}+4}$
• D. $\frac{\sqrt{6}-4}{5}$

Answer 1

The correct option is A $-1\sqrt{3}(\sqrt{2}+\sqrt{3})$

Given that,

$x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}},y=1$

Now,

$\frac{x-y}{x-3y}=\frac{\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}-1}{\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}-3\times 1}$

$\Rightarrow \frac{\sqrt{3}+\sqrt{2}-(\sqrt{3}-\sqrt{2})}{\sqrt{3}+\sqrt{2}-3(\sqrt{3}-\sqrt{2})}$

$\Rightarrow \frac{2\sqrt{3}}{-2\sqrt{3}+2\sqrt{2}}$

$\Rightarrow \frac{\sqrt{3}(\sqrt{2}+\sqrt{3})}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}$

$\Rightarrow \frac{\sqrt{3}(\sqrt{2}+\sqrt{3})}{-1}$

$\Rightarrow -1\sqrt{3}(\sqrt{2}+\sqrt{3})$

Hence, this the answer.