Упростите:

$\frac{z+5}{2z^{2}-2} + \frac{3}{1-z}+\frac{5}{2z+2}}=-\frac{3}{(z-1)(z+1)}$

1) $\frac{z+5}{2z^{2}-2}+\frac{3}{1-z}=\frac{z+5}{2(z^{2}-1)}+\frac{3}{-z+1}=\frac{z+5}{2(z-1)(z+1)}+\frac{3}{-(z-1)}=\frac{z+5-3\cdot2(z+1)}{2(z-1)(z+1)}=\frac{z+5-6(z+1)}{2(z-1)(z+1)}=$ = $\frac{z+5-6z-6}{2(z-1)(z+1)}=\frac{(z-6z)+(5-6)}{2(z-1)(z+1)}=\frac{-5z-1}{2(z-1)(z+1)}= \frac{-(5z+1)}{2(z-1)(z+1)}=-\frac{5z+1}{2(z-1)(z+1)}$

2) $-\frac{5z+1}{2(z-1)(z+1)}+\frac{5}{2z+2}}=\frac{5}{2(z+1)}-\frac{5z+1}{2(z-1)(z+1)}=\frac{5(z-1)-(5z+1)}{2(z-1)(z+1)}=\frac{5z-5-5z-1}{2(z-1)(z+1)}=\frac{(5z-5z)+(-1-5)}{2(z-1)(z+1)}=$ = $\frac{-6}{2(z-1)(z+1)}=\frac{-3}{(z-1)(z+1)}=-\frac{3}{(z-1)(z+1)}$