x+y=1 xy=-12 x(x+y)(x-y)-x(x+y)²的值x+y=1 xy=-12 x(x+y)(x-y)-x(x+y)²的值
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x+y=1 xy=-12 x(x+y)(x-y)-x(x+y)²的值x+y=1 xy=-12 x(x+y)(x-y)-x(x+y)²的值
x+y=1 xy=-12 x(x+y)(x-y)-x(x+y)²的值
x+y=1 xy=-12 x(x+y)(x-y)-x(x+y)²的值
x+y=1 xy=-12 x(x+y)(x-y)-x(x+y)²的值x+y=1 xy=-12 x(x+y)(x-y)-x(x+y)²的值
解 :原式=x(x+y)[x-y-(x+y)]=-2xy(x+y)
x+y=1 以及 xy=-12 代入式子中得
原式=24