设x1x2为方程x²-kx(x-2)+2-k=0的两个实数根.且x1平方+x1x2+x2平方=11/2求k_数学解答

设x1x2为方程x²-kx(x-2)+2-k=0的两个实数根.且x1平方+x1x2+x2平方=11/2求k

人气：272 ℃　时间：2020-02-03 22:21:23

解答

x^2-kx(x-2)+2-k=0(1-k)x^2+2kx+2-k=0x1+x2=-2k/(1-k)=2k/(k-1)x1x2=(2-k)/(1-k)=(k-2)/(k-1)x1^2+x1x2+x2^2=(x1+x2)^2-x1x2=[2k/(k-1)]^2-(k-2)/(k-1)=[4k^2-(k-1)(k-2)]/(k-1)^2=(4k^2-k^2+3k-2)/(k-1)^2=(3k^2+3k...