Question

Желательно как можно быстрей, 100 Баллов...

Answer 1

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

$3)\frac{-4(z+1)}{z(2z-1)(z+2)}:\frac{5}{z^{2} +2z}=\frac{-4(z+1)}{z(2z-1)(z+2)}*\frac{z(z+2)}{5}=-\frac{4(z+1)}{5(2z-1)} \\\\4)-\frac{4(z+1)}{5(2z-1)}-\frac{10z+1}{5-10z}=\frac{4z+4}{5-10z}-\frac{10z+1}{5-10z}=\frac{4z+4-10z-1}{5-10z}=\frac{3-6z}{5-10z}=\\\\=\frac{3(1-2z)}{5(1-2z)}=\frac{3}{5}=0,6\\\\Otvet:\boxed{0,6}$