Question

(K^2十1)x^2一2Kx十(K^2十4)二0跟的情况

Answer 1

（k+1）x-2kx+k+4=0

kx+x-2kx+k+4=0

（x-2kx+k）+kx+4=0

（x-k）+kx+4=0

因（x-k）≥0

kx≥0

所以（x-k）+kx+4≥4

故（k+1）x-2kx+k+4=0无根.