Question

Помогите решить, отдаю все баллы

Answer 1

Ответ:

$\frac{3}{4}$

Объяснение:

$(\frac{z^{2}-2z+4}{4z^{2}-1} \cdot \frac{2z^{2}+z}{z^{3}+8}-\frac{z+2}{2z^{2}-z}):\frac{4}{z^{2}+2z}-\frac{10z+1}{4-8z}=(\frac{(z^{2}-2z+4) \cdot z(2z+1)}{(2z-1)(2z+1) \cdot (z+2)(z^{2}-2z+4)}-$

$-\frac{z+2}{z(2z-1)}) \cdot \frac{z(z+2)}{4}-\frac{10z+1}{4(1-2z)}=(\frac{z}{(2z-1)(z+2)}-\frac{z+2}{z(2z-1)}) \cdot \frac{z(z+2)}{4}+\frac{10z+1}{4(2z-1)}=$

$=\frac{z \cdot z-(z+2) \cdot (z+2)}{z(2z-1)(z+2)} \cdot \frac{z(z+2)}{4}+\frac{10z+1}{4(2z-1)}=\frac{z^{2}-(z^{2}+4z+4)}{z(2z-1)(z+2)} \cdot \frac{z(z+2)}{4}+\frac{10z+1}{4(2z-1)}=$

$=\frac{z^{2}-z^{2}-4z-4}{z(2z-1)(z+2)} \cdot \frac{z(z+2)}{4}+\frac{10z+1}{4(2z-1)}=\frac{-4(z+1) \cdot z(z+2)}{z(2z-1)(z+2) \cdot 4}+\frac{10z+1}{4(2z-1)}=\frac{-4(z+1)}{4(2z-1)}+\frac{10z+1}{4(2z-1)}=$

$=\frac{-4(z+1)+10z+1}{4(2z-1)}=\frac{-4z-4+10z+1}{4(2z-1)}=\frac{6z-3}{4(2z-1)}=\frac{3(2z-1)}{4(2z-1)}=\frac{3}{4};$