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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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人气:335 ℃ 时间:2019-11-24 12:54:51
解答
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=24.752t - t^2 -1 = -(t-1)^2 = 0能解释一下吗?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呃……2(x-y) - (x-y)^2 - 1 = -(x-y-1)^2 是指这步我还是换回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
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