There are two cases, which are not necessarily y=- 1, so they are not the same function.

The absolute value domain of b: y=(x- 1) contains 1, and y = x-1(x > 1)、 1-x(x & lt； 1) There is no 1 in this domain.

So they are not the same function.

C: y = absolute value of x+absolute value of x-1,

For, discuss in two situations.

When x>= 1, this is equal to, when 0

2: f (1/x) = x2+5x = (1/x) (-2)+5 (1/x) (-1) (The solution to this problem is to put forward 65438+ in the original formula.

So f (x) = x (-2)+5x (- 1)

(2x+ 1) in 3:f is equivalent to x in f(x).

So according to 0

4: The domain of 3, y=f(x) is, then the domain of f(2x) [0, 1],

X- 1 as the denominator is not equal to 0. Therefore, the value of x is not 1.

So its domain is [0, 1].

5. Because it is a linear function,

So let y=kx+b,

Substitute 3f (x+1)-2f (x-1) = 2x+17.

3(kx+k+b)-2(kx-k+b)= 2x+ 17

kx+5k+b=2x+ 17

That is k = 2；; b=7

Just a substitute.