暑假作业求助7已知方程组kx^2-x-y+0.5=0 y=k(2x-1) 有两个不同的实数解,{x1 y1 {x2 y2

kx^2-x-y+1/2=0 …………1
y=k(2x-1)………………2
把2带入1
kx^2-x-k(2x-1)+1/2=0
整理得
kx^2-（2k+1）x+1/2+k=0
求△≥0
解(2k+1)^2-4k(1/2+k)≥0
k≥-1/2
y1y2+1/x1+1/x2=3
y1y2+(x1+x2)/x1x2=3
k^2(2x1-1)(2x2-1)+(x1+x2)/x1x2=3
k^2[4x1x2-2(x1+x2)+1]+(x1+x2)/x1x2=3
x1+x2=(2k+1)/k
x1x2=(0.5+k)/k
综上 解得
k=1