Question

y=5x^2-9x+4/x^2-1值域？

Answer 1

y(x²-1)=5x²-9x+4
(y-5)x²+9x-(y+4)=0
x是实数则方程有解
所以判别式81+4(y-5)(y+4)>=0
4y²-4y+1>=0
(2y-1)²>=0
恒成立
所以值域是R

Answer 2

y=5x^2-9x+4/x^2-1
=(5x-4)(x-1)/(x+1)(x-1)
=(5x-4)/(x+1)

再利用反函数法可得:
y(x+1)=5x-4
yx+y=5x-4
yx-5x=-4-y
x=(5x-4)/5-y

可知y的值域为: (负无穷, 5) U (5, 正无穷)