求值域!!!!!!!!!

来源：百度知道 编辑：UC知道 时间：2024/05/22 00:58:54
y=(x^2-1)/(x^2+1)
y=(x^2-x+1)/(x^2+x+1)
y=(3x^2+2)/(x^2+1)

⑴y=(x^2-1)/(x^2+1)=(x^2+1-2)/(x^2+1)=1-2/(x^2+1)
x^2+1的值域为[1,+∞) → -2/(x^2+1)的值域为[-2,0)
→ y=1-2/(x^2+1)的值域为(1,3]

⑵y=(x^2-x+1)/(x^2+x+1)=[(x+1)(x^2-x+1)]/[(x+1)(x^2+x+1)]=(x^3-1)/(x^3+1)=(x^3+1-2)/(x^3+1)=1-2/(x^3+1) （x≠-1时)；x=-1时，y=3
x^3+1 （x≠-1)的值域为R → -2/(x^3+1) （x≠-1)的值域为R → y=1-2/(x^3+1)（x≠-1)的值域为R
→ 综上，y的值域为R

⑶y=(3x^2+2)/(x^2+1)=(3x^2+3-1)/(x^2+1)=3-1/(x^2+1)
x^2+1的值域为[1,+∞) → -1/(x^2+1)的值域为[-1,0)
→ y=3-1/(x^2+1)的值域为(3,4]