Question

rationalize:- 1/√7+√3-√2

Answer 1

this is the right answer

Answer 2

Answer:

$\frac{1}{\sqrt{7}+\sqrt{3}-\sqrt{2}}=\frac{6 \sqrt{3}-8 \sqrt{2}-2 \sqrt{7}+2 \sqrt{42}}{20}$

Solution:

$\frac{1}{\sqrt{7}+\sqrt{3}-\sqrt{2}}$

Simplifying we get,

$\Rightarrow \frac{1}{\sqrt{7}+\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{7}+\sqrt{3}-\sqrt{2}}{\sqrt{7}+\sqrt{3}-\sqrt{2}}$

$\Rightarrow \frac{\sqrt{7}+\sqrt{3}-\sqrt{2}}{\left((\sqrt{7}+\sqrt{3})^{2}\right)-(\sqrt{(2)^{2}})}$

simplifying,

$\Rightarrow \frac{\sqrt{7}+\sqrt{3}-\sqrt{2}}{8+2 \sqrt{21}}$

$\Rightarrow \frac{\sqrt{7}+\sqrt{3}-\sqrt{2}}{8+2 \sqrt{21}} \times \frac{8-2 \sqrt{21}}{8-2 \sqrt{21}}$

$\Rightarrow \frac{2 \sqrt{7}-6 \sqrt{3}+8 \sqrt{2}-2 \sqrt{42}}{8^{2}-(2 \sqrt{21})^{2}}$

On solving we get,

$\Rightarrow \frac{2 \sqrt{7}-6 \sqrt{3}+8 \sqrt{2}-2 \sqrt{42}}{64-84}$

$\Rightarrow \frac{6 \sqrt{3}-8 \sqrt{2}-2 \sqrt{7}+2 \sqrt{42}}{20}$