Question

求y=(x^2-x)/(x^2-x+1)的值域？速回！

Answer 1

原式

分母值域为：[3/4,+∞）

后面的分式值域为：(0,4/3]，加负号以后值域为：[-4/3,0)

整个式子值域为：[-1/3, 1)

Answer 2

解
y=(x²-x)/(x²-x+1)
∴y-1=-1/(x²-x+1)
即1-y=1/(x²-x+1)
易知,该函数定义域为R,
x²-x+1=[x-(1/2)]²+(3/4)≥3/4
∴0＜1/(x²-x+1)≤4/3
即0＜1-y≤4/3
解得:-1/3≤y＜1
即值域为[-1/3, 1)

Answer 3

y=(x^2-x)/(x^2-x+1)=1-1/(x^2-x+1) x^2-x+1=(x-1/2)^2+3/4 x^2-x+1≥3/4
0>-1/(x^2-x+1)≥-4/3 1>1-1/(x^2-x+1)≥1-4/3 =-1/3
y=(x^2-x)/(x^2-x+1)的值域-1/3≤y<1

Answer 4

解法一：y=(x²-x)/(x²-x+1)∴y-1=-1/(x²-x+1)即1-y=1/(x²-x+1)易知,该函数定义域为R,x²-x+1=[x-(1/2)]²+(3/4)≥3/4∴0＜1/(x²-x+1)≤4/3即0＜1-y≤4/3解得:-1/3≤y＜1即值域为[-1/3, 1)
法二：整理成x的二次方程，用△法，更容易懂……