2) When the straight line of A, C and E is * * *, and the value of AC+CE is the smallest, so the intersection BD' provable triangle ABC' and triangle EDC' in AE and C are congruent, then AB:BC'=DE:DC' so.

5:(8-x)= 1:x x=4/3

Therefore, when the CD length is three quarters, the value of AC+CE is the smallest.

3) The graph remains unchanged, but the number changes. According to the formula √ (x 2+4)+√ [( 12-x) 2+9], AB = 3, DE = 2, BD = 12, and CD = X. 。

Similarly, when A, C and E are * * * lines, the value of AC+Ce is the minimum, that is, the minimum value of √ (x 2+4)+√ [(12-x) 2+9].

According to the algorithm in the second question, when x=24/5, the value of AC+Ce is the minimum, that is, the minimum value of √ (x 2+4)+√ [(12-x) 2+9].