已知长方形的周长为20,它的两条邻边长x,y满足2(x-y)-x2+2xy-y2-1=0,求此长方形的 面积
问题描述:
已知长方形的周长为20,它的两条邻边长x,y满足2(x-y)-x2+2xy-y2-1=0,求此长方形的 面积
答
2(x-y)-x2+2xy-y2-1=2(x-y)-(x-y)2-1=-[(x-y)-1]=0,所以x-y-1=0,又2(x+y)=20,联立得x=5.5,y=4.5,所以面积=xy=24.75
答
解:2(x-y)-x^2+2xy-y^2-1=0
2(x-y)-(x^2-2xy+y^2)-1=0
2(x-y)-(x-y)^2-1=0
即(x-y)^2-2(x-y)+1=0
[(x-y)-1]^2=0
所以x-y=1
又因为长方形周长=20
所以2(x+y)=20,x+y=10
当x-y=1,x+y=10时
x=11/2,y=9/2
(11/2)*(9/2)=99/4
希望能帮到你!
答
2(x-y) - (x^2 - 2xy + y^2) - 1 = 0即-2(x-y) + (x-y)^2 +1 = 0(x-y-1)^2 = 0x-y -1 = 0x-y = 1 (1)长方形周长为20,即2x+2y = 20,x+y=10 (2)解(1)(2)组成的二元一次方程组得x=11/2y=9/2 长方形面积=x*y = 99/4...