方程有兩個(gè)根
則判別式=(2k+1)^2-4(k^2+1)=4k-3>=0
k>=3/4
x1>1,x2>1
則(x1-1)(x2-1)>0
且x1+x2>0
x1*x2-(x1+x2)+1
=k^2+1-(2k+1)+1
=k^2-2k+1>0
k不為1
x1+x2=2k+1>0
k>-1/2
綜上,x>3/4且不為1
x2=2x1,
x1+x2=3x1,x1*x2=2x1^2
則2*(x1+x2)^2=9*x1*x2
即2(2k+1)^2=9(k^2+1)
解得k=1或7
再由k的取值范圍知k=7