x(x-1)-(x²-y)=-2,求 二分之(x²+y²)—xy的值
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x(x-1)-(x²-y)=-2,求 二分之(x²+y²)—xy的值
x(x-1)-(x²-y)=-2,求 二分之(x²+y²)—xy的值
x(x-1)-(x²-y)=-2,求 二分之(x²+y²)—xy的值
即x²-x-x²+y=-2
-x+y=-2
x-y=2
所以原式=(x²-2xy+y²)/2
=(x-y)²/2
=2²/2
=2
即x²-x-x²+y=-2
-x+y=-2
x-y=2
所以原式=(x²-2xy+y²)/2
=(x-y)²/2
=2²/2
=2
x(x-1)-(x²-y)=-2化简,-x+y=-2
代数式(x²+y²)/2—xy=(x²+y²—2xy)/2=(x-y)²/2=2