Is the first question wrong?

There are a integers between 5.5 and its reciprocal, and there are b integers between 5.5 and its reciprocal, and | A+ 1 |+B-5 |-| C-2 | = 0. Try to find the value of A+B+C b+c.

The reciprocal of 5.5 is 2/ 1 1, and the reciprocal of 5.5 is-5.5;

From 2/ 1 1 to 5.5, there are 1, 2,3,4,5; a = 5；

The range from -5.5 to 5.5 is -5, -4, -3, -2,-1, 0, 1, 2, 3, 4, 5 respectively; b = 1 1；

|a+ 1|+|b-5|-|c-2|=0

|a+ 1|+|b-5|=|c-2|

5+ 1+ 1 1-5 = | c-2

|c-2|= 12

c 1= 14，c2=- 10

When c= 14

a+b+c = 5+ 1 1+ 14 = 30

When c=- 10

a+b+c = 5+ 1 1- 10 = 6

(2)

Suppose that the distance between a and b is x.

(x-36)÷( 10-8)=(36+36)÷( 12- 10)

x-36 =(36+36)÷( 12- 10)*( 10-8)

x=72+36

x= 108

A: The distance between a:AB and Beijing is108km.

Reference/question/19398058.html? si=3

(3)

Setting: Wang planted eggplant × mu and tomato × mu.

1700 * x+ 1800 *(25-x)= 44000

1700 x+45000- 1800 x = 44000

100x= 1000

x= 10

25- 10= 15 mu

2400 *10+2600 *15 = 63000 (yuan)

A: The Wang family made a net profit of 63,000 yuan.