Question

Rationalize
the
Denominar of 1/√8-√7​

Answer 1

$\Large\mathfrak{\underline{Answer}}$

$\normalsize\textsf{ \sqrt{8}+\sqrt{7} }$

$\Large\textbf{\underline{Given\: :\:}}$

$\normalsize\Longrightarrow\textsf{ \dfrac{1}{\sqrt{8}-\sqrt{7}} }$

$\large\textsf{To rationalise, we will multiply}$

$\large\textsf{the numerator and denominator}$

$\large\textsf{by \sqrt{8}+\sqrt{7} .}$

$\normalsize\Longrightarrow\textsf{ \dfrac{1}{\sqrt{8}-\sqrt{7}}\:\times\:\dfrac{\sqrt{8}+\sqrt{7}}{\sqrt{8}+\sqrt{7}} }$

$\normalsize\Longrightarrow\textsf{ \dfrac{\sqrt{8}+\sqrt{7}}{{\sqrt{8}^2}+{\sqrt{7}^2}} }$

$\normalsize\Longrightarrow\textsf{ \dfrac{\sqrt{8}+\sqrt{7}}{8-7} }$

$\normalsize\Longrightarrow\textsf{ \sqrt{8}+\sqrt{7} }$

$\large\therefore\textbf{The\: Answer\:is\: \sqrt{8}+\sqrt{7} }$

$\Large\textbf{\underline{Note}}$

$\large\odot\:\:\:\textsf{Rationalising is the removal of }$

$\large\textsf{square root from denominator}$

$\large\textsf{by multiplying the fraction with}$

$\large\textsf{rationalising factor.}$

$\large\odot\:\:\:\textsf{Rationalising Factor is a }$

$\large\textsf{number which is multiplied to }$

$\large\textsf{a fraction to remove the square}$

$\large\textsf{root from the denominator of }$

$\large\textsf{the fraction. In the above question}$

$\large\textsf{,the rationalising factor is }$

$\large\textsf{ \sqrt{8}+\sqrt{7} }$

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Answer 2

$\bf \huge \dag \: \: \frac{1}{ \sqrt{8} - \sqrt{7} } \\ \\ \sf = > \frac{1}{ \sqrt{8} - \sqrt{7} } \times \frac{ \sqrt{8} + \sqrt{7} }{ \sqrt{8} + \sqrt{7} } \\ \\ \sf{= > \frac{1( \sqrt{8} + \sqrt{7} )}{ {( \sqrt{8} )}^{2} - {( \sqrt{7} )}^{2} } }\\ \\ \sf = > \frac{ \sqrt{8} + \sqrt{7} }{8 - 7} \: \: = \frac{ \sqrt{8} + \sqrt{7} }{1} \\ \implies\large \boxed{ \frak{ \sqrt{ \red8} + \sqrt{ \red7}} }$

Hence, $\bf{ \sqrt{ \red8} + \sqrt{ \red7} }$ is the rationalised form of $\tt \frac{1}{ \sqrt{8} - \sqrt{7} }$