This problem is solved, and the process needs to be detailed, thank you!

f(x)=(ax+b) (x +1), in (- is the odd function.

f(-x)=-f(x)..or f(0)=0=>b=0]>(ax+b) (x +1)=-(ax+b) (x +1)>(ax+b)=-(ax+b).

b=0 and f(1 2)=(a 2) (1 4+1)=2 5=>a=1 thus f(x)=x (x +1).

If -1 x1 x2 1, then x1-x2 0, -1 x1x2 1, 1-x1x2 0

f(x1)-f(x2)

x1/(x1²+1)-x2/(x2²+1)[(x1-x2)(1-x1x2)] / [(x1²+1)(x2²+1)]＜0

f(x1)＜f(x2)

i.e. f(x) is an increasing function in (-1,1).

1.The odd function has f(-x)=-f(x) so f(-1 2)=-2 5 combined with f(1 2)=2 5 is brought into the function to get a=1 b=0

f(x)=x/x²+1

2.Let x1, x2 (-1x1, whether negative or positive, x2-x1 be greater than 0 (1-x2x1) and greater than 0, so f(x2)-f(x1)>0 is an increasing function.

It's really tiring to type, I haven't done it for a few years, and it seems to be such a solution.

I've solved this problem, and I'll write it again here.

x-m(1-x)=2(1+x) can be obtained by dividing the envy and carrying the mother, and (m-1)x=m+2 is obtained, and if the equation is unsolved, then m-1=0 and m+2≠0; or x=(m+2) (m-1)= 1, from which we get m=1 or -1 2. Brother Zhifu.

Solution: x Xiaochangmeng (1-x)-m (1+x)=2 (1-x)x-m(1-x)=2(1+x).

x+mx-m-2-2x=0

m-1)x=m+2

If m-1=0 is Qiaoqiao m=1, Fang Xunyu Cheng has no solution!

Figure 3 uses 18 sticks. The number of sticks used for the nth figure is (1+2+3+4+......n) 3 sticks.

Because Figure 1 is: 3x1, 1 represents 1 regular triangle.

Figure 2 is: 3x (1+2). 1+2 is the number of small triangles (1 above, 2 below).

Figure 3 is: 3x(1+2+3). Similarly, the number of small triangles is 1 upper, 2nd in the middle, and 3 below.

Figure 4 can be deduced as: 3x(1+2+3+4) and Figure 5 is: 3x(1+2+3+4+5). ...

So: fill in the first blank: 3 (1+2+3)=18.

Second Fill: 3x(1+2+3+......n)

5. The axis of symmetry should be less than or equal to 0 to be satisfied, [0,+ increment.

i.e. x=-a 2 0

a≥06、f(3)=-f(-3)=-7

f(0)=0

So f(0)+f(3)=-7

Question 5: The opening of the quadratic function is upward, the axis of symmetry x=-a 2, increasing at [0,+.

a/2<=0，∴a>=0

Question 6 f(x) is the odd function on r, f(0) = 0

f(3)=-f(-3)=-7

f(0)+f(3)=-7