2(x2-xy)-3(2x2-3xy)-2[x2-(2x2-xy+y2)] 化简一个多项式加上2x2-x3-5-3x4得3x4-5x3-3求这个数的多项式
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2(x2-xy)-3(2x2-3xy)-2[x2-(2x2-xy+y2)]化简一个多项式加上2x2-x3-5-3x4得3x4-5x3-3求这个数的多项式2(x2-xy)-3(2x2-3xy)-2[x2
2(x2-xy)-3(2x2-3xy)-2[x2-(2x2-xy+y2)] 化简一个多项式加上2x2-x3-5-3x4得3x4-5x3-3求这个数的多项式
2(x2-xy)-3(2x2-3xy)-2[x2-(2x2-xy+y2)] 化简
一个多项式加上2x2-x3-5-3x4得3x4-5x3-3求这个数的多项式
2(x2-xy)-3(2x2-3xy)-2[x2-(2x2-xy+y2)] 化简一个多项式加上2x2-x3-5-3x4得3x4-5x3-3求这个数的多项式
2(x²-xy)-3(2x²-3xy)-2[x²-(2x²-xy+y²)]
=2x²-2xy-6x²+9xy-2(x²-2x²+xy-y²)
=-4x²+7xy+2x²-2xy+2y²
=-2x²+5xy+2y²
一个多项式加上2x²-x³-5-3x^4得3x^4-5x³-3,那么可知:
这个多项式=3x^4-5x³-3-(2x²-x³-5-3x^4)
=3x^4-5x³-3-2x²+x³+5+3x^4
=6x^4 -4x³-2x²+2
这个我也需要,^_^
2(x2-xy)-3(2x2-3xy)-2[x2-(2x2-xy+y2)] 化简.
2(x-xy)-3(2x2-3xy)2[x2(2x2-xy+y2)]
(3xy-9x2-2y2-6xy)-(6x2+2xy-3xy-y)
4x2-3y2+2xy-4x2+5x2
已知x2+xy=3,xy+y2=-2,则2x2-xy-3y2=
求证:(x2-xy+y2)3+(x2+xy+y2)3能被2x2+2y2整除
已知x2-3xy=-5,xy+y2=3,求x2-2xy+y2.
若X2+xy-2y2=0,则(x2+3xy+y2)/x2+y2=?
2x2-xy+3y2因式分解
x2 y2+3xy+2 因式分解
x2-3y2=2xy过程
2X2-3XY-Y2分解因式!
4x2+3xy-2y2因式分解
-(x2-2xy-y2)+(5x2-2xy-3y2)-(x2-2xy-y2)+(5x2-2xy-3y2)化简
计算:(xy-x2)÷xy分之x2-2xy+y2
多项式x2-3xy+2y2-2加上( )等于5x2-7xy-3y2
(2x2-3xy+4y2)+(x2+2xy-3y2)化简(要有过程)
解方程组 x2 -2xy -y2 =1 2x2 -5xy -3y2 =0