The domain of the function f x is all real numbers, and if f x 1 f x 1 are both odd functions, then

f(-x+1)=-f(x+1)=0

Let y=x+1, we get f(-y+2)+f(y)=0f(-x-1)=-f(x-1)=0

Let y=x-1, get f(-y-2)+f(y)=0f(-y+2)=f(-y-2), let x=-y-2, get f(x+4)=f(x).

f(x) is a periodic function with a period of 4.

f(x-1) = f(x+3) is an odd function. Pick D

f(x)=f[(x+1)-1]=-f[-(x+1)-1]=-f(-x-2)-f[-(x+3)+1]=f[(x+3)+1]=f(x+4)

f(x+3)=f[(x+3)-4]=f(x-1)=-f(-x-1)=-f[(-x-1)+4]=-f(-x+3)

So: f(x+3) is an odd function.

Choose answer D

The image of f(x+1) is obtained by translating the image of f(x) to the left by 1 unit, and f(x+1) is an odd function, which means that the image of f(x) is symmetrical with respect to the center of the origin after translating 1 unit to the left, so the image of f(x) is symmetrical with respect to the center of point (1,0); In the same way, the image of f(x) is known to be symmetrical with respect to the center of the point (-1,0) since f(x-1) is an odd function; The associative sine function shows that f(x) has an infinite number of symmetry centers, and (3,0) is one of them, so f(x+3) has a symmetry center at the origin, so f(x+3) is an odd function.

Bring x-1 trembling shirt x and -1 round crack x-1 into f(x)+f(x-1 x)=1+x, respectively.

f(x-1 x)+f(-1 x-1)=1+x-1 x,f(-1 x-1)+f(x)=1-1 x-1, the three formulas are connected to the tung cavity, and the solution is f(x)=x3-x2-1 2x(x-1).

f(x+1) defines the auspicious domain as the number of odd letters.

f(-x+1)=-f(x+1)

f(x+1)=-f(-x+1)

If f(x) is an odd function, f(x)=-f(-x) has f(x+1)=-f(-x-1).

Since f(x) is an odd function with the period of 5 as the macro void, then f(x) = f(x+5), and f(-x) = -f(x).

So f(12)=f(7+5)=f(7)=f(2+5)=f(2) and f(-2)=-f(2), so f(2)=-f(-2)=-1 is wide by f(12)=f(2)=-1

f(x-1)=x 2+x+1=(x-1) 2+3(x-1)+3, so f(x)=x 2+3x+3

So f[1 (x-1)]=1 and the mountain (x-1)] 2+3[1 buzhong(x-1)]+3

1/(x-1)^2+3/(x-1)+3

Stick to the definition of "x is an independent variable".

Might as well let x+1=t then x=t-1

Since f(x+1) and f(x-1) are both odd functions, f(x+1) = -f(-x-1) and f(x-1) = -f(-x+1).

This yields f(x+1)=f(t).

f(x-1)=f(t-1-1)=f(t-2) cannot judge the parity of f(x), so ab is wrong.

f(x+2)=f(x+1+1)=f(t+1), so f(x+2)=f(x+1) contradicts option c.

Similarly, f(x+3)=f(x+1+2)=f(t+2)=f(x+1) is an odd function, so d is correct.

Let x+1=t, so f(x+1)=f(t) is an odd function, that is to say, f(x) is an odd function, so a is wrong b is true.

f(x+2)=f(x+1+1)=f(t+1), so f(x+2)=f(x+1), so c is wrong.

f(x+3)=f(x+1+2)=f(t+2)=f(x+1) is an odd function, so d is paired.