Question

rationalize the denominator: 1/1+root2-root3

Answer 1

Question:-

✪Rationalize $\frac{1}{1+\sqrt{2}-\sqrt{3}}$

$\rule{200}{1}$

Answer:-

☞$\frac{\sqrt{2}-\sqrt{3}}{6}$

$\rule{200}{1}$

Proof:-

↝$\frac{1}{1+\sqrt{2}-\sqrt{3}}$

•Multiply both Numerator and Denominator by 1-√2-√3•

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

★Applying a²-b²=(a+b)(a-b) identity★

↝$\frac{1-\sqrt{2}-\sqrt{3}}{(1)^{2}-(\sqrt{2}-\sqrt{3})^{2}}$

↝$\frac{1-\sqrt{2}-\sqrt{3}}{1-(4-9)}$

↝$\frac{1-\sqrt{2}-\sqrt{3}}{1-(-5)}$

↝$\frac{1-\sqrt{2}-\sqrt{3}}{1+5}$

↝$\frac{1-\sqrt{2}-\sqrt{3}}{6}$

$\rule{200}{2}$

Answer 2

...

....

$\tt{\huge{\purple{ ................. }}}$

QUESTION -

rationalize the denominator: 1/1+root2-root3.

SOLUTION -

$\dfrac{1}{ 1 + \sqrt{2} - \sqrt{3} } \\ \\ => \dfrac{1}{ 1 + \sqrt{2} - \sqrt{3} } \times \dfrac{1 + \sqrt{2} + \sqrt{3} }{ 1 + \sqrt{2} + \sqrt{3} } \\ \\=> \dfrac{1 + \sqrt{2} + \sqrt{3} }{ \sqrt{2} - 6 }$

$=> \dfrac{1 + \sqrt{2} + \sqrt{3} }{ \sqrt{2} - 6 } \times \dfrac{ \sqrt{2} + 6 }{ \sqrt{2 } + 6} \\ \\ => \dfrac{1 - \sqrt{2} - \sqrt{3} }{6 }$