Question

初二计算 解方程 不等式组(1)4(y-z）^2-(2y+z)(-z+2y) (2)(x(x^2y^2-xy)-y(x^

Answer 1

1）原式=(4y^2-8yz+4z^2)-(4y^2-z^2)=-8yz+5z^2

2）原式=(x^2y(xy-1)-x^2y(1+xy))/(4x^2y)=(x^2y(xy-1-1-xy))-(4x^2y)=-2/4=-1/2

1）x^2+12x+35-x^2-4x+5=42

8x+40=42

x=2/8

x=1/4

2)2x^2-5x-2x^2+3x-1+7>0

-2x+6>0

xx^2-25-2

12x>-30

x>-2/5

所以 -2/5