Question

rationalize 1 upon root 7 + root 3 minus root 2

Answer 1

Hey mate!!

$\frac{1}{ \sqrt{7} + \sqrt{3} - \sqrt{2} } \\ \\ \frac{1}{ \sqrt{7} + \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{7} + \sqrt{3} + \sqrt{2} }{ \sqrt{7} + \sqrt{3} + \sqrt{2} } \\ \\ \frac{\sqrt{7} + \sqrt{3} + \sqrt{2}}{( \sqrt{7} + \sqrt{3} ) {}^{2} - ( \sqrt{2} ) {}^{2} } \\ \\ \frac{\sqrt{7} + \sqrt{3} + \sqrt{2}}{7 + 2 \sqrt{21} + 3 - 2} \\ \\ \frac{\sqrt{7} + \sqrt{3} + \sqrt{2}}{8 + 2 \sqrt{21} }$

Thanks for the question!!

Answer 2

Given that,

The function is

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

We need to rationalize

Using given function

$=\dfrac{1}{\sqrt{7}+\sqrt{3}-\sqrt{2}}$

Firstly, multiply and divided by $\sqrt{7}+\sqrt{3}+\sqrt{2}$

$=\dfrac{1}{\sqrt{7}+\sqrt{3}-\sqrt{2}}\times\dfrac{\sqrt{7}+\sqrt{3}+\sqrt{2}}{\sqrt{7}+\sqrt{3}+\sqrt{2}}$

$=\dfrac{\sqrt{7}+\sqrt{3}+\sqrt{2}}{(\sqrt{7}+\sqrt{3})^2-(\sqrt{2})^2}$

$=\dfrac{\sqrt{7}+\sqrt{3}+\sqrt{2}}{(\sqrt{7}+\sqrt{3})^2-2}$

$=\dfrac{\sqrt{7}+\sqrt{3}+\sqrt{2}}{(\sqrt{7})^2+(\sqrt{3})^2+2\times\sqrt{7}\times\sqrt{3})-2}$

$=\dfrac{\sqrt{7}+\sqrt{3}+\sqrt{2}}{7+3+2\sqrt{21}-2}$

$=\dfrac{\sqrt{7}+\sqrt{3}+\sqrt{2}}{8+2\sqrt{21}}$

Now, again multiply and divided by $8-2\sqrt{21}$

$=\dfrac{\sqrt{7}+\sqrt{3}+\sqrt{2}\times(8-2\sqrt{21})}{8+2\sqrt{21}\times8-2\sqrt{21}}$

$=\dfrac{8\sqrt{7}+8\sqrt{3}+8\sqrt{2}-2\sqrt{147}-2\sqrt{63}-2\sqrt{42}}{8^2-(2\sqrt{21})^2}$

$=\dfrac{8\sqrt{7}+8\sqrt{3}+8\sqrt{2}-2\sqrt{147}-2\sqrt{63}-2\sqrt{42}}{64-4\times21}$

$=\dfrac{8\sqrt{7}+8\sqrt{3}+8\sqrt{2}-2\sqrt{147}-2\sqrt{63}-2\sqrt{42}}{64-84}$

$=\dfrac{8\sqrt{7}+8\sqrt{3}+8\sqrt{2}-2\sqrt{147}-2\sqrt{63}-2\sqrt{42}}{-20}$

$=\dfrac{8\sqrt{7}+8\sqrt{3}+8\sqrt{2}-14\sqrt{3}-6\sqrt{7}-2\sqrt{42}}{-20}$

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

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

Hence, This is required solution.