Question

If x=
$1 \div 3 + 2 \sqrt{2}$

then find the value of
$x - 1 \div x$

​

Answer 1

Step-by-step explanation:

$x = \frac{1}{3 + 2 \sqrt{2} } \\ \\ \\ = > x - \frac{1}{x} = > \frac{1}{3 + 2 \sqrt{2} } - \frac{1}{ \frac{1 }{ 3 + 2 \sqrt{2} } } \\ \\ \\ = > \frac{1}{3 + 2 \sqrt{2} } - 3 + 2 \sqrt{2} \\ \\ \\ = > \frac{1 - {(3 + 2 \sqrt{2)} }^{2} }{3 + 2 \sqrt{2} } \\ \\ \\ = > \frac{1 - (9 + 8 + 12 \sqrt{2} )}{3 + 2 \sqrt{2} } \\ \\ \\ = > \frac{1 - 17 - 12 \sqrt{2} }{3 + 2 \sqrt{2} } \\ \\ \\ = > \frac{ - 15 - 12 \sqrt{2} }{3 + 2 \sqrt{2} } \\ \\ \\ = > \frac{ - 15 - 1 2\times1.41 }{3 + 2 \times 1.41} \\ \\ \\ = > \frac{ - 15 - 16.92}{3 + 2.82} \\ \\ \\ = > \frac{ - 31.92}{5.82} = > - 8.35$

please mark me brainlist ✔️