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Solution: Set up to rent X trains of type A and Y trains of type B, according to the relationship in the question, there are:

x+y=16

18x+16y≥266

10x+11y≥169

Draw the above three equations to establish the coordinate axis, and you will get a region.

Because the fuel cost = 1500x+1200y, so let 1500x+1200y be the smallest, that is, on the line of x+y=16, and the integer solution closest to the origin in the region is the solution, it can be found that when taking 18x+16y=269 and 10x+11y=169 near the intersection point, it should be the closest to the origin in the three regions, and the nearest integer interprets x=5, y=11, so the solution is to rent 5 cars of the first type, 11 cars of type B, the minimum is 5 1500 + 11 1200 = 20700 yuan.

Solution: (1) Set up to rent X trucks of type A, and rent (16-x) trucks of type B, according to the title, 18x+16(16x) 266

10x+11(16−x)≥169②

By , x 5, by , x 7, so, 5 x 7, x is a positive integer, x = 5 or 6 or 7, therefore, there are 3 car rental schemes: scheme 1: group A type of trucks 5, type B 11 trucks.

Plan 2: Group 6 type A trucks and 10 type B trucks. Option 3:

There are 7 trucks of type A and 9 trucks of type B. (2) Method 1: From (1), rent X trucks of type A and rent trucks of type B for (16-x), and the total fuel cost of the two trucks is Y yuan.

According to the title, y=1500x+1200(16-x), =300x+19200, 300 0, when x=5, y has a minimum value, y min=300 5+19200=20700 yuan;

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After conversion you get (a+b) 2 (a-b) 2=2, then (a+b) (a-b)=+sqrt(2).

Proof: (1).

Prove the monotonicity of a function by definition].

Take x1, x2 r, and x1 x2

then f(x1) f(x2) a [2 (2 x1 1)] a [2 (2 x2 1)].

2(2 x1 2 x2)] [(2 x1 1)(2 x2 1)] y 2 x increments on ( and x1 x2 2 x1 2 x2

2^x1)－(2^x2)＜0

(2 x1 1) (2 x2 1) 0

f(x1)－f(x2)＜0

i.e. f(x1) f(x2).

f(x) is an increment function on (.

2) f(x) is an odd function, then f(0) a [2 (2 0 1)] a 1 0

A 1 is tested and f(x) is an odd function when a 1.

3) f(t 2+2) + f (t 2-tk) > 0 is constant and immediately has f(t 2+2) >-f(t 2-tk) = f(tk-t 2) and t 2 is obtained by the increase function t 2+2>tk-t 2

2t^2-tk+2>0

Then there is discriminant = k 2-16<0

i.e. get -4

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