Question

已知X,Y是一元二次方程4KX平方-4KX+K+1=0的两个实根

Answer 1

x+y=1,xy=(k+1)/4k

1.原式=2x*2+2y*2-5xy

=2[(x+y)*2-2xy]-5xy

=2(x+y)*2-9xy

=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=(x*2+y*2)/xy-2

=[(x+y)*2-2xy]/xy-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