Question

已知X1,X2是一元二次方程 4KX平方-4KX+k+1=0的两个实数根.

Answer 1

2^x2 2的x2次方?应该是2x2吧?

利用韦达定理

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