Question

求(x^2-x)/(x^2+x)的值域

不好意思，我打错了，是(x^2-1)/(x^2+1)的值域，抱歉！

Answer 1

y=(x²-x)/(x²+x)
=x(x-1)/x(x+1)
=(x-1)/(x+1)
=[(x+1)-2]/(x+1)
=1-2/(x+1)
x≠0所以y≠-1
x≠-1所以y≠1
如果是
y=(x²-1)/(x²+1)
=[(x²+1)-2]/(x²+1)
=1-2/(x²+1)
x²+1>=1,则-2<=-2/(x²+1)<0
所以-1<=y<1

Answer 2

(x^2-1)/(x^2+1)
=1-2/（x^2+1） (x^2+1>=1，推出0<2/（x^2+1<=2)
所以原式的值域为[-1,1）

Answer 3

y=x(x-1)/x(x+1)
=(x-1)/(x+1)
=(x+1-2)/(x+1)
=(x+1)/(x+1)-2/(x+1)
=1-2/(x+1)

2/(x+1)≠0
所以1-2/(x+1)≠1

又分母m(m+1)≠0
所以m≠0
则2/(x+1)≠2
1-2/(x+1)≠-1

所以值域(-∞,-1)∪(-1,1)∪(1,+∞)