[x²-(x+y)(x-y)][x²-2xy-(x-y)²],其中x=-1,y=-2 化简求值 急急急 要过程
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[x²-(x+y)(x-y)][x²-2xy-(x-y)²],
=（x^2-x^2+y^2）(x^2-2xy-x^2+2xy-y^2)
=y^2(-y^2)
=-y^4
=-(-2)^4
=-16
=（x^2-x^2+y^2）(x^2-2xy-x^2+2xy-y^2)
=y^2(-y^2)
=-y^4
=-(-2)^4
=-16
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(x+y)(x-y)=x²-y²
(x-y)²=x²-2xy+y²
[x²-(x+y)(x-y)][x²-2xy-(x-y)²]
=[x²-(x²-y²)][x²-2xy-(x²-2xy+y²)]
=y²×(-y²)
=-y⁴
=-(-2)⁴
=-16
(x-y)²=x²-2xy+y²
[x²-(x+y)(x-y)][x²-2xy-(x-y)²]
=[x²-(x²-y²)][x²-2xy-(x²-2xy+y²)]
=y²×(-y²)
=-y⁴
=-(-2)⁴
=-16
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