题目内容：

已知y=5x²+x+1/x+1,若x是实数,求y值的范围.

优质解答

y=5x²+x+1/x+1
5x²+x+1-yx-y=0
5x²+(1-y)x+1-y=0
x有解
Δ = (1-y)²-20(1-y)>=0
(1-y-20)(1-y)>=0
(y-1)(y+19)>=0
y>=1或y