A set of desks and chairs, the desk is 33 yuan more expensive than the chair. It is known that the price ratio of the desk and chair is 4:7. How much are the desks and chairs?

A set of desks and chairs, the desk is 33 yuan more expensive than the chair. It is known that the price ratio of the desk and chair is 4:7. How much are the desks and chairs?


Because their ratio is 4:7 And they know that there is a difference of 33 yuan between them, we can solve the problem of difference times
7-4=3
33 * 3 = 11 yuan
11 * 4 = 44 yuan
11 * 7 = 77 yuan
A: the desk is 77 yuan and the chair is 44 yuan



The price of a set of desks and chairs is 120 yuan, of which the price of chairs is 57 yuan of that of desks. How much is the price of chairs?


120 × 55 + 7, = 120 × 512, = 50 yuan; answer: the price of chair is 50 yuan



1. The price of a set of desks and chairs is 120 yuan, of which the price of chairs is 2 / 7 cheaper than that of desks. How much are the prices of desks and chairs?
2. A road, the first day 1 / 4, the second time 2 / 5 more than the first time, there are still 28 kilometers left. How many kilometers is the total length of this road?
3. A train goes from a place to B place. After 9 / 13 of the whole journey, it goes at a speed of 120 km / h for 3 hours to get to B place. How many km is the distance from a place to B place?


1. Table: 120 ÷ (1-2 / 7 + 1) = 70 yuan; Chair: 120-70 = 50 yuan
2. 1-1 / 4-2 / 5 = 7 / 20 28 △ 7 / 20 = 80 (km)
3. 120 × 3 = 360 (km) 360 △ (1-9 / 13) = 1170 (km)



The price of a set of desks and chairs is 48 yuan, of which the price of chairs is 57 yuan of that of desks. How much is the price of desks?


A: the price of the desk is 28 yuan



Number set a satisfies the following conditions: if a ∈ a, then 11 − a ∈ a (a ≠ 1) (1) if 2 ∈ a, try to find out all other elements in a (2) design a number to belong to a, and then find out all other elements in a (3). What can you learn from the above solution? And boldly prove the "truth" you found


(1) If 2 ∈ a, then 11 − 2 = − 1 ∈ a, 11 + 1 = 12 ∈ a, 11 − 12 = 2 ∈ a, that is, all other elements in a are - 1, 12. (2) if 3 ∈ a, then 11 − 3 = − 12 ∈ a, 11 + 12 = 23 ∈ a, 11 − 23 = 3 ∈ a, that is, all other elements in a are − 12, 23. (3) there are only three elements a, 11 − a, a − 1a in a, and the result of three numbers is - 1 And 11 − a ≠ 1), then 11 − 11 − a = a − 1a ∈ a, and a − 1a ≠ 1, and then 11 − a − 1A = a ∈ a, ∵ a ≠ 11 − a (if a = 11 − a, that is a2-a + 1 = 0, then the equation has no solution) ≠ 11 − a ≠ a − 1a, there are only three elements a, 11 − a, a − 1a in ≠ a, and the score of three numbers is - 1



Let a be a set of numbers satisfying a ∈ a, then 1 / (1-A) ∈ a 1.. if 2 ∈ a, it is proved that a contains at least three elements 2. It is proved that if a ∈ a, then 1 - (1 / a) ∈ a


If a ∈ a, then 1 / (1-A) ∈ a, then 1 / [1-1 / (1-A)] = (A-1) / a ∈ a, then 1 / [1 - (A-1) / a] = a ∈ a, so when the above three expressions are not equal to each other, this set contains and only contains three elements: A, 1 / (1-A), (A-1) / A, respectively. It is proved that when a = 2, the three elements are 2, - 1,1 / 2, so there are three elements, it is proved that



If a ∈ a, then 1 + A1 − a ∈ a (a ≠ 1). If 13 ∈ a, find other elements in the set


∵ 13 ∈ a, ∵ 1 + 131 − 13 = 2 ∈ a, ∵ 1 + 21 − 2 = − 3 ∈ a, that is, 1 − 31 + 3 = − 12 ∈ a, so 1 − 121 + 12 = 13 ∈ a, and the elements cycle and repeat. Therefore, when 13 ∈ a, the other elements in the set are 2, - 3, - 12



If a ∈ a, then 1 + A1 − a ∈ a (a ≠ 1). If 13 ∈ a, find other elements in the set


∵ 13 ∈ a, ∵ 1 + 131 − 13 = 2 ∈ a, ∵ 1 + 21 − 2 = − 3 ∈ a, that is, 1 − 31 + 3 = − 12 ∈ a, so 1 − 121 + 12 = 13 ∈ a, and the elements cycle and repeat. Therefore, when 13 ∈ a, the other elements in the set are 2, - 3, - 12



How to build a new sequence from the general term formula of a sequence? Can you give an example?


An = n + 1, (number 2,3,4,...)
An=an+5
Namely
An = n + 6 (number 7,8,9,...)



Second order construction formula for general term of sequence
Known sequence {an}, A1 = 5, A2 = 2, an = 2an-1 + 3an-2 (n ≥ 3), find the general term formula. Note | lowercase letter is the subscript, because I won't type that. Please. If it's good, I will add points. The whole process of construction


an=2a(n-1)+3a(n-2)an-3a(n-1)=a(n-1)+3a(n-2)an-3a(n-1)=(-1)(a(n-1)-3a(n-2))bn=an-3a(n-1),b2=a2-3a1=-1bn=(-1)b(n-1)bn=(-1)^(n-2)b2bn=(-1)^(n-1)an-3a(n-1)=(-1)^(n-1)an+a(n+1)=3(a(n-1)+a(n-2))an+a(n-1)=3^...