• 计算：（x+y)*x2/x2-y2+y2/y-x
• x2-2xy+y2-1/x-y-1
• 已知两圆C1：x²＋y²＝1，C2：（x－2）²＋（y－2）²＝5，求经过点P（0,1）且被两圆截
• x2+y2 xy+x+y-1

计算：（x+y)*x2/x2-y2+y2/y-x

（x+y)*x^2/(x^2-y^2)+y^2/(y-x)

=（x+y)*x^2/(x-y)(x+y)-y^2/(x-y)

=x^2/(x-y)-y^2/(x-y)

=(x^2-y^2)/(x-y)

=(x-y)(x+y)/(x-y)

=x+y

x2-2xy+y2-1/x-y-1

x2-2xy+y2-1/x-y-1

=【(x-y)²-1】/(x-y-1)

=(x-y-1)(x-y+1)/(x-y-1)

=x-y+1

已知两圆C1：x²＋y²＝1，C2：（x－2）²＋（y－2）²＝5，求经过点P（0,1）且被两圆截

设直线的方程为：y=kx+1

C1：x²＋y²＝1，圆心(0,0)到直线的距离d1=1/根号(k^2+1),半径r1=1

弦长=2*根号[k^2/(k^2+1)]

C2：（x－2）²＋（y－2）²＝5，，圆心(2,2)到直线的距离d2=|2k-1|/根号(k^2+1),半径r2=根号5

弦长=2*根号[（k^2+4k+4)/(k^2+1)]

弦长相等

2*根号[k^2/(k^2+1)]=2*根号[（k^2+4k+4)/(k^2+1)]

k=-1

直线的方程

y=-x+1

x2+y2 xy+x+y-1

(x2+y2)-(xy+x+y-1)

=(1/2)*[(x^2-2xy+y^2)+(x^2-2x+1)+(y^2-2y+1)]

=(1/2)*[(x-y)^2+(x-1)^2+(y-1)^2]

因为(x-y)^2≥0,(x-1)^2≥0,(y-1)^2≥0

(三项都取=号,有解x=y=1)

所以

(x2+y2)-(xy+x+y-1)≥0

x^2+y^2≥xy+x+y-1