y=x^2+2x/x^2+2x+3求函数值域

y=(x^2+2x+1-1)/(x^2+2x+1+2)
=[(x+1)^2-1]/[(x+1)^2+2]
另(x+1)^2=t则t≥0
y=(t-1)/(t+2)
=(t+2-3)/(t+2)
=1-3/(t+2)
t≥0则t+2≥2
0＜1/(t+2)≤1/2
-3/2≤-3/(t+2)＜0
-1/2≤1-3/(t+2)＜1
所以函数的值域为[-1/2,1)

是y=(x^2+2x)/(x^2+2x+3)吧？
y=(x^2+2x)/(x^2+2x+3)
=1 -3/(x^2+2x+3)
=1 -3/((x+1)^2+2)
(x+1)^2+2>=2
00>-3/((x+1)^2+2)>=-3/2
1>1 -3/((x+1)^2+2)>=-1/2
函数值域为[-1/2，1)

y=(x^2+2x+3-3)/(x^2+2x+3)
=1-3/(x^2+2x+3)
=1-3/[(x+1)^2+2]
因为(x+1)^2+2>=2,所以0

是y=(x²+2x)/(x²+2x+3)吧！
y=(x²+2x)/(x²+2x+3)=1-3/(x²+2x+3)=1-3/[(x+1)²+2]，
由(x+1)²+2≥2可知-3/2≤-3/[(x+1)²+2]所以-1/2≤y