Question

$\dfrac { {(3-2√2)}^{2} }{ {(3)}^{2} - {(2√2)}^{2} } = ?$ ​

Answer 1

Answer :

$\frac{(3 - 2 \sqrt{2} ) {}^{2} }{(3) {}^{2} - (2 \sqrt{2}) {}^{2} } \\ \\ \frac{(3) {}^{2} +( 2 \sqrt{2}){}^{2} - 2 \times3 \times 2 \sqrt{2} }{9 - 8} \\ \\ \frac{9 + 8 - 12 \sqrt{2} }{1} \\ \\ \boxed{ \bf 17 - 12 \sqrt{2}}$

Thanks!