(x²-2xy+y²)+(-2x+2y)+分解(x²-2xy+y²)+(-2x+2y)+1
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(x²-2xy+y²)+(-2x+2y)+分解(x²-2xy+y²)+(-2x+2y)+1
(x²-2xy+y²)+(-2x+2y)+分解
(x²-2xy+y²)+(-2x+2y)+1
(x²-2xy+y²)+(-2x+2y)+分解(x²-2xy+y²)+(-2x+2y)+1
(x²-2xy+y²)+(-2x+2y)+1
=(x-y)²-2(x-y)+1
=(x-y-1)²
(x²-2xy+y²)+(-2x+2y)
=(x-y)²-2(x-y)
=(x-y)(x-y-2)
(x-y)(x-y-2)
x²-2xy+y²-9
若x²+xy-2y²=0,则x²+3xy+y²/x²+y²
x²+xy-2y²=0则x²+3xy+y²/x²+y²
因式分解 x²-2xy+y²-z²
计算:4xy²-3x²y-{3x²y-[2xy²-4x²y+2(x²y-2xy²)]急
x²-4y²/x²+2xy+y²÷x+2y/2x²+2xy
计算x²-x/x²×x/1-x和x²-4y²/x²+2xy+y²÷x+2y/x²+xy
x²-y²+6x+9分之x²+y²-9-2xy
因式分解 x²-XY-2y²-X-Y
x²+xy-2y²-x+7y-6
(xy-x²)÷[(x²-2xy+y²)÷xy]*[(x-y)÷x²]
6x²-xy-15y²=0 求x²-2xy+4y²/2x²-3xy-3y²告诉下
6x²-xy-15y²=0 求2x²-3xy-3y²分之x²-2xy+4y²
谢谢高手解答x²-xy=2,-xy+y²=5求x²-2xy+ y²,x²- y²
因式分解2x(x-y)^4 - x²(x-y)² + xy(y-x)²RT
已知:x²+xy=3,xy+y²=-2 求:2x²-xy-3y²
(x+2y)²-x²-2xy
因式分解X²-3Xy+2y²+X-2