Question

求值域 y=x²-1/x²+1 y=x²-x/x²-x+1

求过程

Answer 1

y=(x²-1)/(x²+1) = (x²+1-2)/(x²+1) = 1 - 2/(x²+1)
x²+1≥1
0＜2/(x²+1)≤2
-2≤-2/(x²+1)＜0

-1≤1-2/(x²+1)＜1
值域【-1，1）

y=(x²-x)/(x²-x+1) = (x²-x+1-1)/(x²-x+1) = 1 - 1/[(x-1/2)²+3/4]
(x-1/2)²+3/4≥3/4
0＜1/[(x-1/2)²+3/4]≤4/3
-4/3≤-1/[(x-1/2)²+3/4]＜0
-1/3≤1-1/[(x-1/2)²+3/4]＜1
值域【-1/3，1)

追问

问一下第一个有一步去分母为什么是1-2/（x²+1）

追答

y =(x²-1)/(x²+1)
= [ (x²+1)-2)]/(x²+1)
= (x²+1)/(x²+1) - 2/(x²+1)
= 1 - 2/(x²+1)