Known: three consecutive odd numbers, their sum of squares is 251, find these three odd numbers

Known: three consecutive odd numbers, their sum of squares is 251, find these three odd numbers

Let these three odd numbers be n-2, N, N + 2 in turn, where n is a natural number, then n > 2, then according to the problem, we can get the following equation: (n-2) 2 + N2 + (n + 2) 2 = 251, 3N2 = 243, N2 = 81, ∧ n = 9 or n = - 9, when n = 9, n-2 = 7, N + 2 = 11; when n = - 9, n-2 = - 11, N + 2 = - 7; answer: these three consecutive odd numbers are 7

Cut a square iron plate with a side length of 9.5 decimeters into two right triangle small iron plates with right angle sides of 4.5 decimeters and 1 decimeter respectively. How many pieces can be cut finally?

9.5÷4.5＝2……………… 0.5 decimeter
9.5÷1＝9……………… 0.5 decimeter
9 × 2 × 2 = 36 pieces

Science in daily life!

For example, toothpaste can clean teeth, because there are enamel and some active ingredients in toothpaste

Given the absolute value of x-y-2 + the square of (2xy-1) = 0, find the square of X (x + y) (X-Y) - x (Y-X)

It is easy to know that x-y-2 = 0, 2xy-1 = 0
x-y=2,2xy=1
The original formula = x (x + y) (X-Y) - x (X-Y) & sup2;
=x(x-y)(x+y-x+y)
=2xy(x-y)
=1×2=2

a. B is a non-zero vector. The equivalent condition for proving that a and B are perpendicular is | a + B | = | a-b|

(1) A + B | = | A-B | A is perpendicular to B, so AB = 0 (a + b) ^ 2 = a ^ 2 + B ^ 2 + 2Ab = a ^ 2 + B ^ 2-2ab = (a-b) ^ 2, so | a + B | = | A-B | (2) a is perpendicular to B | a + B | = | A-B | so (a + b) ^ 2 = (a-b) ^ 2, so AB = 0, that is, a is perpendicular to B

It is known that D and E are the points along the long line of BC and Ba on the equilateral triangle ABC, and BD = AE. It is proved that CE = De

If DF = BC is made on BD extension line, then BF = be, angle FBE = 60 degrees, and bef is equilateral triangle, then be = EF, BC = DF, angle FBE = angle EFB are deduced, then triangle BCE is equal to triangle EFD, and CE = De is obtained

Given that a sequence an is an arithmetic sequence, and A1 is not equal to 0, Sn is the sum of the first n terms of the sequence, find limnan/ Sn.limSn +Sn-1/Sn+Sn-1

1. Sn = (a1 + an) n / 2, so Nan / Sn = 2An / (a1 + an) = 2 [A1 + (n-1) D] / [2A1 + (n-1) D] divided by (n-1) = 2 [A1 / (n-1) + D] / [2A1 / (n-1) + D] n-1 tends to infinity, so 1 / (n-1) tends to 0, so limit = 2D / D = 22, original formula = [n (n + 1) / 2 + (n + 1) (n + 2) / 2] / [n (n + 1) / 2 + n (n-1) / 2] =

Four coplanar examples of space vector
Four points a (2,1, - 3), B (- 2,3, - 4), C (3,0,1) are known,
D (1,4, m), if a, B, C and D are coplanar, then M = ()
A、-7 B、-22 C、19 D、5

B. - 22 fourth order determinant = 0
|1 2 1 -3|
|1 -2 3 -4|
|1 3 0 1|
|The solution is m = - 22

As shown in the figure, in RT △ ABC, CD is the height on the hypotenuse AB, point m is on CD, DH ⊥ BM and the extension line of AC intersects at point e. the following are proved: (1) △ AED ∽ CBM; (2) AE · cm = AC · CD

The following: (1) the following proof: (1) ABC is a right triangle, right triangle, right triangle, right triangle, right triangle, right triangle, right triangle, and right triangle, the \\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\cbm, AE: Ad =CB: cm, ∵ AE · cm = ad · CB, ∵ ABC is a right triangle, CD is the high on AB, ∵ ACD ∽ CBD, ∵ AC: ad = CB: CD, ∵ AC · CD = ad · CB, ∵ AE · cm = AC · CD

It's better not to have an equation. Even if you want an equation, you have to have only one unknown number
1. The master and the apprentice process a batch of parts. The apprentice's efficiency is three fifths of that of the master. The master first makes a total of one third of the parts, and then the apprentice completes them. They share 52 minutes. How many minutes did the apprentice do?
2. Xiao Ming went to his grandfather's house by bike on Sunday. After three fourths of the whole ride, the car broke down and he had to push it. It took 33 minutes to get there. It is known that the speed of the cart is two fifths of that of the bicycle. How many minutes did it take?

If the master's speed is V, the apprentice's speed is 3 / 5V, and the total number of parts can be regarded as 1. If the master does one third of the total, the time is (1 / 3) / v = 1 / (3V). If the apprentice does two thirds of the total, the time is (2 / 3) / (3 / 5V) = 10 / (9V). The total time is 1 / (3V) + 10 / (9V) = 13 / (9V) = 52