(x+3y-2z)(x-3y-2z)-(x+3y-2z)^2 要过程
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(x+3y-2z)(x-3y-2z)-(x+3y-2z)^2 要过程
还有一道:
填空:(x-y+1)(x+y-5)=(x-y-2+_____)(x+y-2-_____)=[(x-2)-(____)]*[(x-2)+(____)]=(____)^2-(____)^2=_________
还有一道:
填空:(x-y+1)(x+y-5)=(x-y-2+_____)(x+y-2-_____)=[(x-2)-(____)]*[(x-2)+(____)]=(____)^2-(____)^2=_________
(x+3y-2z)(x-3y-2z)-(x+3y-2z)^2
=(x+3y-2z)[(x-3y-2z)-(x+3y-2z)]
=(x+3y-2z)(-6y)
=-6xy-18y²+12yz
再问: (x+3y-2z)^2的平方呢
再答: 提取公因式(x+3y-2z) 另一个因式是 [(x-3y-2z)-(x+3y-2z)] (x-y+1)(x+y-5)=(x-y-2+3)(x+y-2-3)=[(x-2)-(y-3)]*[(x-2)+(y-3)]=(x-2)^2-(y-3)^2=x²-4x-y²+6y-5
=(x+3y-2z)[(x-3y-2z)-(x+3y-2z)]
=(x+3y-2z)(-6y)
=-6xy-18y²+12yz
再问: (x+3y-2z)^2的平方呢
再答: 提取公因式(x+3y-2z) 另一个因式是 [(x-3y-2z)-(x+3y-2z)] (x-y+1)(x+y-5)=(x-y-2+3)(x+y-2-3)=[(x-2)-(y-3)]*[(x-2)+(y-3)]=(x-2)^2-(y-3)^2=x²-4x-y²+6y-5
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
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