(1-XY)(1-XY)+(X+Y-2)(x+y-2XY)因式分解 A>B>C,求证AB^2+B^2C+C^2A
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(1-XY)(1-XY)+(X+Y-2)(x+y-2XY)因式分解A>B>C,求证AB^2+B^2C+C^2A(1-XY)(1-XY)+(X+Y-2)(x+y-2XY)因式分解A>B>C,求证AB^2
(1-XY)(1-XY)+(X+Y-2)(x+y-2XY)因式分解 A>B>C,求证AB^2+B^2C+C^2A
(1-XY)(1-XY)+(X+Y-2)(x+y-2XY)因式分解 A>B>C,求证AB^2+B^2C+C^2A
(1-XY)(1-XY)+(X+Y-2)(x+y-2XY)因式分解 A>B>C,求证AB^2+B^2C+C^2A
原式=（x+y-xy-1）^2
设x+y=a,xy=b
则原式=（1-b）^2+(a-2)(a-2b)=1-2b+b^2+a^2-2ab-2a+4b=a^2-2(b+1)a+(b+1)^2=(a-b-1)^2=(x+y-xy-1)^2
第二题题目有问题,不是等式怎么证明啊?
同意第二题的题有问题，A>B>C,没有说明是否大于0，这个可能是你要假设的条件。
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