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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^2+x^3y)/4x^2y
(1)(x+7)(x+5)-(x-1)(x+5)=42(2)x(2x-5)>(x-1)(2x-1)-7
(x+1)(x+3)+8x>(x+5)(x-5)-2
人气:313 ℃ 时间:2020-05-26 08:29:22
解答
1)原式=(4y^2-8yz+4z^2)-(4y^2-z^2)=-8yz+5z^22)原式=(x^2y(xy-1)-x^2y(1+xy))/(4x^2y)=(x^2y(xy-1-1-xy))-(4x^2y)=-2/4=-1/21)x^2+12x+35-x^2-4x+5=428x+40=42x=2/8 x=1/42)2x^2-5x...
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