Question

Rationalize:-
$\orange{ \tt{ \dfrac{ \sqrt{2} }{ \orange{ \sqrt{2} + \sqrt{3} - \sqrt{5} } }}}$
-Rendom question 201-
$\large \bold \red{@WildCat7083}$
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Answer 1

$\large{\pmb{\sf{\underline{Required\: solution...}}}}$

The question says that we need to rationalize the denominator of a fraction. The fraction is mentioned below:

$\pmb{ \sf{ \dfrac{ \sqrt{2} }{ \sqrt{2} + \sqrt{3} - \sqrt{5} } }}$

Explanation:

We need to reduce the fraction as much as possible until the denominator gets rationalised.

How to rationalize the denominator? By reducing it! First, We will find the R.F of the denominator and multiply it with it! Then, we will use laws of exponents only if needed and solve it until the denominator is rationalised!

Step by step explanation:

R.F of √2 + √3 - √5 is √2 + √3 + √5,

${\hookrightarrow \: \pmb{ \sf{ \dfrac{ \sqrt{2} }{ \sqrt{2} + \sqrt{3} - \sqrt{5} } \times \dfrac{\sqrt{2} + \sqrt{3} + \sqrt{5}}{\sqrt{2} + \sqrt{3} + \sqrt{5}} }} }$

Multiplying,

${\hookrightarrow \: \pmb{ \sf{ \dfrac{ \sqrt{2}(\sqrt{2} + \sqrt{3} + \sqrt{5})}{ {(\sqrt{2}} + { \sqrt{3} )}^{2} - {( \sqrt{5})}^{2} } }}}$

${\hookrightarrow \: \pmb{ \sf{ \dfrac{ \sqrt{2}(\sqrt{2} + \sqrt{3} + \sqrt{5})}{ {2 + 3 +2 \sqrt{6} } - {( \sqrt{5})}^{2} } }}}$

${\hookrightarrow \: \pmb{ \sf{ \dfrac{ \sqrt{2}(\sqrt{2} + \sqrt{3} + \sqrt{5})}{ {5 +2 \sqrt{6} } - {5} } }}}$

${\hookrightarrow \: \pmb{ \sf{ \dfrac{ \sqrt{2}(\sqrt{2} + \sqrt{3} + \sqrt{5})}{ {2 \sqrt{6} } } }}}$

${\hookrightarrow \: \pmb{ \sf{ \dfrac{ \sqrt{2}(\sqrt{2} + \sqrt{3} + \sqrt{5})}{ {2 \sqrt{6} } } \times \dfrac{ \sqrt{6} }{ \sqrt{6} } }}}$

${\hookrightarrow \: \pmb{ \sf{ \dfrac{ \sqrt{12}(\sqrt{2} + \sqrt{3} + \sqrt{5})}{ {12} } { } }}}$

${\hookrightarrow \: \pmb{ \sf{ \dfrac{ 2 \sqrt{3} (\sqrt{2} + \sqrt{3} + \sqrt{5})}{ {12} } { } }}}$

${\hookrightarrow \: \pmb{ \sf{ \dfrac{ \sqrt{3} (\sqrt{2} +\sqrt{3} + \sqrt{5})}{ {6} } { } }}}$

Henceforth, Denominator of the fraction is rationalised.

_______________________

Additional information:

• Rational numbers: The numbers which can be written in p/q form. where q ≠ 0 i.e., q is not equal to zero. For example - 7/8, 9/8, 567/738.
• Irrational numbers: Irrational numbers are the totally opposite to rational numbers. They cannot be expressed in the form of p/q. The best examples for irrational numbers is e and π.

Answer 2

EXPLANATION:-

$\sf\;\frac{\sqrt{2} }{\sqrt{2}+\sqrt{3}-\sqrt{5} }$

Rationalising factor of √2+√3-√5 is √2+√3+√5 .

So we will multiply it by it's R.F

$\\\sf\;\frac{\sqrt{2} }{\sqrt{2}+\sqrt{3}-\sqrt{5} }\times\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{\sqrt{2}+\sqrt{3}+\sqrt{5}} \\\\==>\frac{\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{5})}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}\\\\==>\frac{\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{5})}{2+3+2\sqrt{6}-(\sqrt{5})^2}\\\\==>\frac{\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{5})}{5+2\sqrt{6}-5}}\\\\==>\frac{\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{5})}{5+2\sqrt{6}+5}}\\\\==>\frac{\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{5})}{2\sqrt{6}}\\\\Rationalising\;again$

$\sf\;\===>\frac{\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{5})}{2\sqrt{6}}\\\\==>\frac{\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{5})}{2\sqrt{6}}\times\frac{\sqrt{6}}{\sqrt{6}}\\\\==>\frac{\sqrt{12}(\sqrt{2}+\sqrt{3}+\sqrt{5}}{12} \\\\==>\frac{2\sqrt{3}(\sqrt{2}+\sqrt{3}+\sqrt{5}}{12}\\\\==>\frac{\sqrt{3}(\sqrt{2}+\sqrt{3}+\sqrt{5}}{6}$

Hence the denominator is rationalized

$\red\rule{300}{5}$