Question

please tell me solution of this question

We have to rationalise the denominator ​

Answer 1

Answer:-

$\\ \tt\mapsto \dfrac{1 + \sqrt{2} }{2 - \sqrt{2} } .$

$\\ = \tt\frac{1 + \sqrt{2} }{2 - \sqrt{2} } \times \frac{2 + \sqrt{2}}{2 + \sqrt{2} }$

$\\ = \tt \dfrac{(1 + \sqrt{2})(2 + \sqrt{2}) }{(2 + \sqrt{2})( 2 - \sqrt{2} ) } .$

$\\ = \tt \frac{2 + \sqrt{2} + 2 \sqrt{2} + 2}{(2) {}^{2} - ( \sqrt{2}) {}^{2} }$

As (a+b) (a-b) = a²-b².

$\\ = \tt \frac{ 4 + 3 \sqrt{2} }{4 - 2}$

$\\ \large{\red{ \bf= \boxed{ \blue{ \bf\frac{4 + 3 \sqrt{2} }{2} .}}}}$

Answer 2

Answer:

$\frac{4 + 3 \sqrt{2} }{2}$

Step-by-step explanation:

Question:

$\frac{1 + \sqrt{2} }{2 - \sqrt{2} }$

Solution:

$\frac{1 + \sqrt{2} }{2 - \sqrt{2} }$

==> $\frac{{1}\mathrm{{+}}\sqrt{2}}{{2}\mathrm{{-}}\sqrt{2}}$×$\frac{{2}\mathrm{{+}}\sqrt{2}}{{2}\mathrm{{+}}\sqrt{2}}$

==> $\frac{{\mathrm{(}}{1}\mathrm{{+}}\sqrt{2}{\mathrm{)(}}{2}\mathrm{{+}}\sqrt{2}}{{2}^{2}\mathrm{{-}}{\mathrm{(}}\sqrt{2{\mathrm{)}}^{2}}}$

==> $\frac{{1}{\mathrm{(}}{2}\mathrm{{+}}\sqrt{2}{\mathrm{)}}\mathrm{{+}}\sqrt{2}{\mathrm{(}}{2}\mathrm{{+}}\sqrt{2}{\mathrm{)}}}{{4}\mathrm{{-}}{2}}$

==> $\frac{{2}\mathrm{{+}}\sqrt{2}\mathrm{{+}}{2}\sqrt{2}\mathrm{{+}}{\mathrm{(}}\sqrt{2}{\mathrm{)}}^{2}}{2}$

==> $\frac{{4}\mathrm{{+}}{3}\sqrt{2}}{2}$ Ans.

${}^{i \: hope \: its \: help \: you.}$