Question

数学题求值域。

Answer 1

你好，请采纳！

本题适宜用有界函数法：

从原表达式中解x^2,再利用x^2≥0,得出y的范围；
......................................................................
y=(x^2-1)/(x^2+1)
x^2-1=yx^2+y
x^2(1-y)=(1+y)
x^2=(1+y)/(1-y)≥0
(y-1)(y+1)≤0(y≠1)
-1≤y<1
原函数的值域为：[-1,1)

当然也有普通的方法

1.
y = (xx+1-2)/(xx+1) = 1 - 2/(xx+1)
0 < 2/(xx+1) <= 2
=>
-1<=y<1

2.
y可以取到0，
当y不为0时
y = 1/(1 + 2/(xx - 1))
2/(xx-1) <= -2
或
2/(xx-1) > 0
=>
-1=<y<0
y<1
综上-1<=y<1

Answer 2

Answer 3

原式可以等价为y+x*x*y=x*x-1;
x*x*y-x*x=-y-1;
x*x(y-1)=-y-1;
因为y不等于1；（当y=1时，x无解。）
x*x=(-y-1)/(y-1);
因为x*x>=0;
所以 (-y-1)/(y-1)>=0;
(y+1)/(y-1)<=0;
所以-1<=y<=1;由前面可知y不等于1；
所以-1<=y<1。

Answer 4

追答

求采纳，么么哒

Answer 5

求导