On the first day, I climbed 3 meters during the day and dropped 2 meters at night. The formula is 3-2= 1, and the cumulative rising height is 1 m.

The next day, I climbed 3 meters during the day and dropped 2 meters at night. The formula is 3-2+3-2=2, and the cumulative rising height is 2m.

On the third day, climb 3 meters during the day and drop 2 meters at night. The formula is 3-2+3-2+3-2=3, and the cumulative rising height is 3m.

On the fourth day, climb 3 meters during the day and drop 2 meters at night. The formula is 3-2+3-2+3-2+3-2=4, and the cumulative rising height is 4m.

On the fifth day, I climbed 3 meters during the day, 4+3=7 meters, and I have climbed to the wellhead.

So there was a well 7 meters deep. A snail climbed up from the bottom of the well, climbing 3 meters during the day and falling 2 meters at night. This snail can climb out of the well in five days.

Extended data:

This kind of problem belongs to engineering problems in mathematical application.

1. Basic quantitative relation

1, work efficiency × time = total workload

2, work efficiency = total workload ÷ working hours

3, working hours = total workload ÷ work efficiency

Second, the basic characteristics

Let the total workload be "1" and the work efficiency be = 1/ times.

Three: basic methods

Arithmetic method, proportional method, equation method.

Four: basic ideas

Do it together, do it together.

Another solution is: suppose it takes x days for a frog to climb out of a 6-meter-deep well, then according to the meaning of the question,

(3-2)*(X- 1)+3=7, and X=5 can be obtained by solving the equation.

4-3 is the actual climbing height of snails every day. On the X day, the snail can climb out of the wellhead by climbing 3 meters, so X- 1 represents the cumulative climbing height before the X day.

So there was a well 7 meters deep. A snail climbed up from the bottom of the well, climbing 3 meters during the day and falling 2 meters at night. This snail can climb out of the well in five days.