Question

长方形周长20，两条邻边长x，y满足2（x-y）-x的二次方+2xy-y的二次方-1=0,求此长方形的面积

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Answer 1

设x-y = t
则2t - t^2 -1 = -(t-1)^2 = 0
t = 1
所以x-y=1
因为2x+2y=20
所以x = 5.5, y = 4.5
所以面积是S=xy=24.75

追问

2t - t^2 -1 = -(t-1)^2 = 0
能解释一下吗？

追答

2(x-y) - x^2 + 2xy - y^2 -1 = 2(x-y) - (x-y)^2 - 1 = -(x-y-1)^2 = 0
得x-y = 1

追问

呃……2(x-y) - (x-y)^2 - 1 = -(x-y-1)^2 是指这步

追答

我还是换回t吧

2t - t^2 -1 = - (t^2 - 2t + 1) 一眼看出来是个完全平方公式 = -(t-1)^2

Answer 2

长方形周长20，有：x+y=10
由：2(x-y)-x²+2xy-y²-1 = 2(x-y)-(x-y)²-1 = -(x-y-1)² =0
可得：x-y-1=0
于是可解得：x=5.5， y=4.5
面积是xy=24.75