方程由兩實(shí)數(shù)根,則判別式
(-4k)^2-4*4k*(k+1)≥0
-16k≥0 ∴ k≤0
x1,x2,為方程4kx²-4kx+k+1=0的兩實(shí)數(shù)根
x1+x2=1,x1*x2=(k+1)/(4k)
(1) (2x1-x2)(x1-2x2)=-1.5
2x1^2+2x2^2-5x1x2=-1.5
2(x1+x2)^2-9x1x2=-1.5
2-9*(k+1)/(4k)=-1.5
k=-0.2 符合題意
當(dāng)k=-0.2 時(shí),(2x1-x2)(x1-2x2)=-1.5成立
（2.x1/x2+x2/x1-2
=(x1^2+x2^2-2x1x2)/(x1x2)
=[(x1+x2)^2-4x1x2]/(x1x2)
=[1-4*(k+1)/(4k)]/(k+1)/(4k)
=-4/(k+1)
使x1/x2+x2/x1-2的值為整數(shù)
所以 -4/(k+1)為整數(shù)
由于 k≤0 所以 k只能是0,-2,-3,-5
3.k=-2:
x1/x2+x2/x1-2=-4/(-1)=4
x1/x2+x2/x1-6=0
f+1/f-6=0
f^2-6f+1=0
f=(6(+/-)4根號(hào)2)/2=3(+/-)2根號(hào)2