Question

全忘了！！！请大神帮我解答一下！！！！！！谢谢！！！！要有过程！！！！！一定采纳！！！

Answer 1

x1+x2=1,x1x2=(k+1)/4k
1、原式=2x1*2+2x2*2-5x1x2
=2[(x1+x2)*2-2x1x2]-5x1x2
=2(x1+x2)*2-9x1x2
=2-9(k+1)/4k=-3/2
k=9/5
b*2-4ac=16k*2-4k(k+1)=12k*2-4k=128.88>0
k=9/5

2、原式=(x1*2+x2*2)/x1x2-2
=[(x1+x2)*2-2x1x2]/x1x2-2
={1-[2(k+1)/4k]}/[(k+1)/4k]-2
=(2k-2)/(k+1)-2
=-4/(k+1)
要使此式为整数,则k+1能被-4整除
k+1=1,-1,2,-2,4,-4
k=0,-2,1,-3,3,-5

追答

求采纳