Solving the first year math problem analysis process Thank you

Suppose you have a sum of money for investment, there are 3 options for you to choose from, the returns of these 3 plans are as follows, plan 1, return 40 yuan per day, plan 2, return 10 yuan on the first day, and then return 10 yuan every day compared with the previous day, plan 3, return yuan on the first day, and double the return of each day after the previous day, may I ask, which investment plan to choose has the greatest return.

Solution: Use n to represent the number of business days, and use y to form the total return (yuan) in n days, then:

Scheme 1: y = 40n

Scheme 2: y = 10n + 5n (n-1) = 5n + 5n = 5n (n + 1) Scheme 3: y =

Comparison between Option 1 and Option 2: When n=7 days is the total return of the two, the total return of the two is equal; When n > 7 days, y > y

Comparison between Option 2 and Option 3: when n=10 days, y = 550 yuan; y = yuan;

When n=14 days, y = 1050 yuan; y = yuan;

Which plan pays off the most, you say!?

To be honest, I think this problem seems to be less conditional, but the flower explained by the image should be the third type. The first is a straight line, the second is an oblique straight line, and the third is a curve on the opening image, and the return should be the area of the image, which day is counted, because the third type is changing faster and faster, so I think it is the third type. You can try it out by drawing it yourself.

I don't know that either.

The WJ on the second floor is already very detailed, why are you still asking for help.

If there is a place where you don't understand, you can continue to ask questions.

From the left you can push out the right, from the right you can push out the left, and the horizontal line is filled with the conditions.

Sufficient necessities, a b = a, only a is in b, or a = b. i.e. a is included in the definition of b.

The three questions are actually of the same type.

Question 1: If you change the x-y wheel to a-b, you can do it.

Way of thinking: a-b can be seen as [(a-b) 1 2] 2, to prove greater than or equal to, only to prove that the left side of the inequality minus the right side of the inequality results are greater than or equal to zero.

1/(a-b)+1/(b-c)-1/(a-b)^1/2*1/(b-c))^1/2

1 (a-b) 1 2] 2+[1 (b-c) 1 2] 2)-1 (a-b) 1 2*1 (b-c)) 1 cavity surplus 2

1/(a-b)^1/2+1/(b-c)^1/2]^2≥0

The original formula is proven. If question 2 is not an equal sign, it can be done if it is greater than or equal to the sign.

y/x+9x/y-2√9=[√y/x)]^2+[√9x/y)]^2-2√[（y/x）*(9x/y)]=y/x)-√9x/y)]^2≥0

i.e. y x+9x wax round xun y 2 9

The original formula is proven. 3 questions.

x＾2+y＾2-2xy=(x-y)^2≥0

i.e. x 2 + y 2 2xy

c "If he has a number on the top right (set to m) and a number on the bottom right (set to n), then he should divide the class n by the class of m by the class of (n-m).

Example 1,. 1) If two of the three principals are from middle school A, then we can think that the two who have selected middle school A only need to choose one of the remaining four, that is, c upper right 1 lower right 4 = 4, and the method of selecting three from five middle schools is c upper right 3 lower right 6 = 10, then the probability is compared with the two: 4 20 = 1 5

2) Method 1: Divided into two situations1and one was from A; 2.All three are from other schools.

Method 2: Directly subtract the probability of coming from the same school in 1 question, that is, 1-1 5=4 5 5 If there is a problem hi me, if you are satisfied, hope).

The fastest way to solve this problem is to directly bring special values into it, the first A1B minus 1C1D minus 1, which adds up to 0, so A is wrong, and the second ABCD is equal to minus 1, so B is wrong, C answer can not be satisfied, D answer ABCD is equal to 1, and the sum is equal to 4, so choose D

Multiple choice questions, it is recommended that the special value method is quickly answered, assuming that they are all 1, then the value is 4, which is obviously D correct.

Normal solution ideas: ABC three-time timing: 4; When ABC is two positive and one negative: 0; When abc is positive and negative: 0; ABC triple loss: -4

It's too easy to pull! The formula is well memorized, that is, two perfectly square formulas, upstairs correct solution.