Question

Найти значение выражения. (смотри картинку ниже).

Answer 1

$\displaystyle \frac{4z}{z^2-9}- \frac{2}{z-3}- \sqrt{2} = \frac{4z}{(z-3)(z+3)} - \frac{2(z+3)}{(z-3)(z+3)}-\sqrt{2} =\\ \\ \\ = \frac{4z-2z-6}{(z-3)(z+3)} -\sqrt{2}= \frac{2z-6}{(z-3)(z+3)}-\sqrt{2}= \frac{2(z-3)}{(z-3)(z+3)}-\sqrt{2}=\\ \\ \\ = \frac{2}{z+3} -\sqrt{2}$

Если $z=\sqrt{2}-3$, то

$\displaystyle \frac{2}{\sqrt{2}-3+3} -\sqrt{2} =\sqrt{2}-\sqrt{2}=0$

Answer 2

$\frac{4z}{(z+3)(z-3)} - \frac{2(z+3)}{(z-3)(z+3)} - \sqrt{2} = \frac{2(z-3)}{(z-3)(z+3)}$ $- \sqrt{2}$  =
$\frac{2- \sqrt{2} ( \sqrt{2} -3) - 3 \sqrt{2} }{( \sqrt{2} -3) +3}$
$= \frac{2-2+3 \sqrt{2} - 3 \sqrt{2} }{ \sqrt{2} } = 0$