Question

root2-1 by root2+1=a+b root2​

Answer 1

$\sf\large \frac{ \sqrt{2} - 1 }{ \sqrt{2} + 1 } = a + b \sqrt{2}$ rationalise the denominator

$\rm \large{ \frac{ \sqrt{2} - 1 }{ \sqrt{2} + 1 } \times \frac{ \sqrt{2} - 1}{ \sqrt{2} - 1} = a + b \sqrt{2} }$

$\rm \large \frac{ {(\sqrt{2} - 1) }^{2} }{ {(\sqrt{2} )}^{2} - {1}^{2} } = a + b \sqrt{2}$

$\rm \large \frac{ {{(\sqrt{2})}^{2} -2( \sqrt{2} )(1) + 1}^{2} }{ 2 - 1} = a + b \sqrt{2}$

$\rm \large \frac{ {2 + 1 -2\sqrt{2} } }{ 2 - 1} = a + b \sqrt{2}$

$\rm \large { {3 -2\sqrt{2} } } = a + b \sqrt{2}$

$\star{ \rm{\: a = 3}} \\ \star{ \: \sf{b = - 2 }}$