Analysis: Using the solution set of inequality and Vieta's theorem to get two relations, and then simplifying them with known conditions to get the value of A. 。

Answer:

Solution: because the solution set of inequality x2-2ax-8a 2 < 0 (a > 0) about x is (x 1, x2),

So x 1+x2=2a…①, x 1? X2 =-8a 2 … ②, while x2-x 1= 15…③,

So choose a.

Tianjin

When a= 1, f(x)=x|x|+x,

∵f(x+a) 0 is a constant on R, and it is changed into △ < 0, thus the inequality about A is obtained, and the range of A is obtained.

Answer:

Solution: Because the inequality X 2-AX+2A > 0 is constant on r 。

∴△ = (-a) 2-8a < 0, then the solution is 0 < a < 8.

So the answer is: (0,8)

Sichuan Province

Analysis: If the property of even function is f(|x+2|)=f(x+2), then f (x+2) < 5 can be changed into f (|x+2|) < 5, and the inequality can be represented by a known expression. Find the range of | x+2 | first, and then find the range of x.

Solution: Solution: Because f(x) is an even function, f(|x+2|)=f(x+2).

Then f (x+2) < 5 can be changed into f (| x+2 |) < 5, that is | x | x+2 | 2-4 | x+2 | < 5, (| x+2 |+ 1) (| x+2 |-5) < 0,

So | x+2 | < 5, the solution is -7 < x < 3,

So the solution set of inequality f (x+2) < 5 is (-7,3).

So the answer is: (-7,3).