Question

Rationalize denominator 1 by 5 + under root 2

Answer 1

\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{3}

Step-by-step explanation:

Rationalising\: the \: \\denominator \: of \: \frac{1}{\sqrt{5}+\sqrt{2}}

Multiply numerator and denominator by (√5-√2), we get

=\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}

=\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}

/* By algebraic identity:

(a+b)(a-b)=a²-b²*/

=\frac{\sqrt{5}-\sqrt{2}}{5-2}

=\frac{\sqrt{5}-\sqrt{2}}{3}

/* Denominator rationalised */

Therefore,

\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{3}

Answer 2

${\huge{\pink{\underline{\underline{\purple{\mathbb{\bold{\mathcal{AnSwEr..}}}}}}}}}$

${\large{\blue{\bold{\underline{IDs\:Used:-}}}}}$

=》 (a+b)(a-b) = a²-b².

${\large{\blue{\bold{\underline{Given:-}}}}}$

$\implies \frac{1}{5 + \sqrt{2} } \\ \\ = \frac{1}{5 + \sqrt{2} } \times \frac{5 - \sqrt{2} }{5 - \sqrt{2} } \\ \\ = \frac{5 - \sqrt{2} }{( {5})^{2} - ( \sqrt{2}) {}^{2}} \\ \\ = \frac{5 - \sqrt{2} }{25 - 2} \\ \\ = \frac{ 5 - \sqrt{2} }{23}$

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