跪求解答一道二次方程的数学题已知x1,x2是一元两次方程4kx²-4kx+k+1的两个实数根 （1）是否存在实

(1) x1,x2为方程4kx²-4kx+k+1=0的两个根
则x1+x2=1, x1x2=(k+1)/(4k)
2(2x1-x2)(x1-2x2)+3=0 => 2[2x1²+2x2²-5x1x2]=2[2(x1+x2)²-9x1x2]+3=0
=> 2[2-9(k+1)/(4k)]+3=0 => k=9/5
∴存在k=9/5使方程成立
(2) x1/x2+x2/x1=(x1²+x2²)/(x1x2)=[(x1+x2)²-4x1x2]/(x1x2)
=(1-4x1x2)/(x1x2)=1/(x1x2)-4=4k/(k+1)-4=-4/(k+1)
欲使上述式子为整数,则-4/(k+1)必为整数
∴k+1必为4的约数,易知,k取整数时,有 k+1=±1,±2,±4；
∴满足条件的k的取值有：k=-5,-3,-2,0,1,3

(1)由题意得Δ=-16k≥0
∴k≤0
∵a=4k≠0
∴k≠0
∴k＜0
由根与系数的关系得
x1+x2=1
x1·x2=(k+1)/4k
如果存在实数k，使2（2x1-x2）（x1-2x2）+3=0成立
4x1²-10x1x2+4x2²+3=0
4(x1+x2)²-18x1x2+3=...