Ответы
Ответ:
3/4
Объяснение:
[(z²-2z+4) / (4z²-1) * (2z²+z)/(z³+8) - (z+2)/(2z²-z)]:4/(z²+2z)-(10z+1)/(4-8z) =
=[(z²-2z+4)/(2z-1)(2z+1) *z(2z+1)/(z³+8) -(z+2)/(z(2z-1)]:4/(z(z+2) -(10z+1)/4(1-2z)=
=[(z²-2z+4)/(2z-1) *z/(z³+8)-(z+2)]:4/z(z+2-(10z+1)/4(1-2z)=
=[(z²-2z+4z)*z/(2z-1)(z³+8) - (z+2)/z(2z-1)] :4/z(z+2) - (10z+1)/4(1-2z)=
=[(z²-2z+4)z/(2z-1)(z+2)(z²-2z+4) -(z+2)/z(2z-1) ] : 4/z(x+2) -(10z+1)/4(1-2z)=
=[(z/(2z-1)(z+2) - (z+2)/z(2z-1)]:4/z(z+2) - (10z+1)/4(1-2z)=
=(z²-(z+2)²)/z(2z-1)(z+2) *z(x+2)/4 - (10z+1)/4(1-2z)=
=(z²-(z+2)²)/(2z-1) *1/4 - (10z+1)/4(1-2z)=
=(z²-(z+2)²)/4(2z-1) - (10z+1)/4(1-2z)=
=(z²-(x+2)²/4(2z-1) + (10z+1)/4(2z-1)=
=(z²-(z+2)²+10z+1)/4(2z-1)=
=(z²-(z^2+4z+4)+10+1)/4(2z-1)=
=(z²-z²-4z-4+10z+1)/4(2z-1)=
=(6z-3/4(2z-1)=3(2z-1)/4(2z-10= 3/4