x^2-xy-2y^2-x+5y-2
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x^2-xy-2y^2-x+5y-2x^2-xy-2y^2-x+5y-2x^2-xy-2y^2-x+5y-2首先,X^2-XY+2Y^2=(X-2Y)(X+Y)所以,设x^2-xy-2y^2-x+5y
x^2-xy-2y^2-x+5y-2
x^2-xy-2y^2-x+5y-2
x^2-xy-2y^2-x+5y-2
首先,X^2-XY+2Y^2=(X-2Y)(X+Y)
所以,设x^2-xy-2y^2-x+5y-2可分解成
(X-2Y+A)(X+Y+B)
则展开有X的一次项=A+B=-1
Y的一次项有A-2B=5
连列成方程,解出A=1 B=-2
所以有原式可以分解为:(X-2Y+1)(X+Y-2)
x^2-xy-2y^2-x+5y-2
=(x-2y)(x+y)-x+5y-2
=(x-2y)(x+y)+(x+y)-2x+4y-2
=(x+y)(x-2y+1)-2(x-2y+1)
=(x+y-2)(x-2y+1)
(-x-y)(x^2+xy+y^2)
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2*(x-y)²(y-x)(xy)
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x^2-xy-2y^2-x+5y-2
-x^5y-xy+2x^3y
2X^+XY-Y^-4X+5Y-6
2x^+xy-y^-4x+5y-6=?