Question

решить у меня получается два, но не правильно пишет

Answer 1

$1-\frac{3}{3-z}=\frac{18}{z^{2}-9 }+\frac{8}{3+z} \\\\1+\frac{3}{z-3}=\frac{18}{z^{2}-9 }+\frac{8}{z+3} \\\\1+\frac{3}{z-3}-\frac{18}{(z-3)(z+3) }-\frac{8}{z+3}=0\\\\\frac{z^{2}-9+3*(z+3)-18-8*(z-3) }{(z-3)(z+3)}=0\\\\\frac{z^{2}-9+3z+9-18-8z+24 }{(z-3)(z+3)}=0\\\\\frac{z^{2}-5z+6 }{(z-3)(z+3)}=0\\\\\frac{(z-3)(z-2)}{(z-3)(z+3)}=0\\\\\left \{ {{(z-3)(z-2)=0} \atop {z\neq3;z\neq-3}} \right.\Rightarrow z=2$

Проверка :

$1-\frac{3}{3-2}=\frac{18}{2^{2}-9 } +\frac{8}{3+2}\\\\1-3=\frac{18}{4-9}+\frac{8}{5}\\\\-2=\frac{18}{-5}+1,6\\\\-2=-3,6+1,6\\\\-2=-2-verno\\\\Otvet:\boxed{2}$

Answer 2

$2$

Объяснение:

ОДЗ:

$z\neq-3, \quad z\neq 3;$

$1-\frac{3}{3-z}=\frac{18}{z^{2}-9}+\frac{8}{3+z};$

$\frac{3-z-3}{3-z}=\frac{18}{(z-3)(z+3)}+\frac{8}{z+3};$

$\frac{-z}{3-z}=\frac{18+8(z-3)}{(z-3)(z+3)};$

$\frac{z}{z-3}=\frac{18+8z-24}{(z-3)(z+3)};$

$z(z+3)=8z-6;$

$z^{2}+3z-8z+6=0;$

$z^{2}-5z+6=0;$

$\left \{ {{z_{1}+z_{2}=5} \atop {z_{1}\cdot z_{2}=6}} \right. \Leftrightarrow \left \{ {{z_{1}=2} \atop {z_{2}=3}} \right. ;$

Корень z₂ не удовлетворяет ОДЗ.