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1 x 2 y 2xy

(xy-1)²+(x+y-2)(x+y-2xy) =(xy)²-2xy+1+(x+y)²-2(x+y)-2xy(x+y)+4xy =(xy+1))²+(x+y)²-2(x+y)(xy+1) =(xy+1-x-y)²

x²+2xy+y²=1 (1) 2x²-y=2 (2) 由(1)得 (x+y)²=1 ∴x+y=1 x+y=-1 因此可得到两个新的方程组 x+y=1 x+y=-1 2x²-y=2 2x²-y=2 解这两个方程组得 x=1 x=-3/2 x=-1 x=1/2 y=0 y=5/2 y=0 y=-3/2

化为: [(1+x²)y'-2xy]/(1+x²)²=1 [y/(1+x²)]'=1 积分得: y/(1+x²)=x+C 得y=(1+x²)(x+C)

2xy-x²-y²+1 =-(x²-2xy+y²-1) =-[(x-y)²-1] =-(x-y+1)(x-y-1)

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

(x+y-2xy)(x+y-2)+(xy-1)² =(x+y)²-(2xy+2)(x+y)+4xy+x²y²-2xy+1 =(x+y)²-2(xy+1)(x+y)+(x²y²+2xy+1) =(x+y)²-2(xy+1)(x+y)+(xy+1)² =[(x+y)-(xy+1)]² =(x+y-xy-1)²

ax²+bxy+cy²+dx+ey+f=0 这里a=1,b=-2,c=-1 所以△=b²-4ac>0 所以是双曲线

x2 – 2xy + 3y2 – 2=(x-y)^2-2+2y=(x-y)^2-2(1+y^2)=(x-y)^2-2(1+y)(1-y)因为 y= x – 1 代入所以式子变为(x-x+1)^2-2(1+x-1)(1-x+1)=1-2x(2-x)=2x^2-4x+1

由x,y满足2≤y≤4-x,x≥1,画出可行域如图所示.则A(2,2),B(1,3).x2+y2+2x?2y+2xy?x+y?1=(x+1)2+(y?1)2(x+1)(y?1)=x+1y?1+y?1x+1,令k=y?1x+1,则k表示可行域内的任意点Q(x,y)与点P(-1,1)的斜率.而kPA=2?12?(?1)=13,kPB=3?11?(...

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