一道超级难的题目的说。
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原式=1[(x-y)-(x²-y²)]-1/[1-(x²+2xy+y²)]
=1/[(x-y)-(x+y)(x-y)]-1/[(1-(x+y)²)]
=1/[(x-y)(1-x-y)]-1/[(1+x+y)(1-x-y)]
=[(1+x+y)-(x-y)]/[(x-y)(1-x-y)(1+x+y)]
=(1+x+y-x+y)/(x-y)(1-x-y)(1+x+y)]
=(1+2y)/[(x-y)(1-x-y)(1+x+y)]
=1/[(x-y)-(x+y)(x-y)]-1/[(1-(x+y)²)]
=1/[(x-y)(1-x-y)]-1/[(1+x+y)(1-x-y)]
=[(1+x+y)-(x-y)]/[(x-y)(1-x-y)(1+x+y)]
=(1+x+y-x+y)/(x-y)(1-x-y)(1+x+y)]
=(1+2y)/[(x-y)(1-x-y)(1+x+y)]
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分子(1+2y)
分母(x-y)(1-x-y)(1+x+y)
分母(x-y)(1-x-y)(1+x+y)
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分母不等于零。两个等式,解x,y
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是计算不是方程
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