Question

please tell me fast it's urgent ​

Answer 1

$\huge \boxed { \sf \blue{a = 3 . b = - 2}}$

Step-by-step explanation:

Q.

$\sf \: find \: the \: value \: of \: a \: and \: b \\ \sf \frac{ \sqrt{2} - 1}{ \sqrt{2} + 1 } = a + b \sqrt{2}$

Solution -

$\sf \: first \: we \: will \: rationalise \: the \: denominator \\ \\ \sf = > \frac{ \sqrt{2} - 1}{ \sqrt{2} + 1} \\ \\ = > \sf \frac{ \sqrt{2} - 1 }{ \sqrt{2} + 1} \times \frac{ \sqrt{2} - 1}{ \sqrt{2} - 1} \\ \\ = > \sf\frac{ {( \sqrt{2} - 1)}^{2} }{( \sqrt{2} {)}^{2} - {1}^{2} } \\ \sf (by \: the \: formula \: (a + b)(a - b) = {a}^{2} - {b}^{2} ) \\ \\ = > \sf \frac{( \sqrt{2} {)}^{2} - 2( \sqrt{2} )(1) + {1}^{2} }{2 - 1} \\ ( \sf \: by \: the \: formula \: ( {a - b)}^{2} = {a}^{2} - 2ab + {b}^{2} ) \\ \\ = > \sf \frac{2 - 2 \sqrt{2} + 1}{1} \\ \\ = > \sf2 + 1 - 2 \sqrt{2} \\ \\ = > \sf3 - 2 \sqrt{2}$

Hence a = 3 & b = -2

hope it helps.