Question

y=(3x^2+3x+1)/（x^2+x+1）求用判别法求值域.

Answer 1

y(x²+x+1)=3x²+3x+1

(y-3)x²+(y-3)x+y-1=0.

y=3时,方程不成立,故y≠3.

△=(y-3)²-4(y-3)(y-1)≥0

即(y-3)(1-3y)≥0

所以1/3≤y≤3.但y≠3,所以1/3≤y