The mathematical formula of grade one to grade six in primary school

The mathematical formula of grade one to grade six in primary school

Number of copies per copy = total number of copies ／ number of copies = total number of copies ＋ number of copies = 1 times × multiple = several times several times ＋ 1 times = Times several times ＋ times = 1 times speed × time = distance ＋ speed = time distance ＋ time = speed unit price × quantity = total price of total price ／ unit price = total quantity ＋ quantity = unit price work efficiency × working time = total amount of work Total work ÷ work efficiency = working time total work ÷ working time = work efficiency Plus + addend = sum - one adder = another addend minus - subtraction = subtraction difference = subtraction difference + subtraction = subtraction factor × factor = product △ one factor = divisor of another factor △ divisor = quotient divisor × divisor = divisor Primary school mathematics figure calculation formula square C perimeter s area a side length perimeter = side length × 4 C = 4A area = side length × side length s = a × a cube V volume a edge length surface area = edge length × edge length × 6 s surface = a × a × 6 volume = edge length × edge length v = a × a × a rectangle C perimeter? S area a side perimeter = (length + width) × 2 C = 2 (a + b) area = length × width s = AB 4 cuboid V volume s area? A length? B width H height (1) surface area (length × width + length × height + width × height) × 2 s = 2 (AB + ah + BH) (2) volume = length × width × height v = ABH triangle s area a bottom h height area = bottom × height △ 2 s = ah △ 2 triangle height = area × 2 △ bottom triangle bottom = area × 2 △ parallelogram s area a bottom h height area = bottom × height s = ah trapezoid s area a upper bottom B bottom h height area = (upper bottom + bottom) × height △ 2 s = (a + b) × h △ 2 circle S area C perimeter Π d = diameter r = radius (1) perimeter = diameter ×Π = 2 ×Π? Radius C = Πd = 2 Π R (2) area = radius × radius ×Πcylinder V volume? H height? S; bottom area? R bottom radius C bottom perimeter (1) side area = bottom perimeter × height (2) surface area = side area + bottom product × 2 (3) volume = bottom area × height (4) volume = side area ﹤ 2 × radius

Clear organization

The number of copies per copy = the total number of copies △ the number of copies per copy = the total number of copies △ the number of copies = 2.1 times × multiple = several times several times △ 1 times = several times several times △ 1 times = Times several times × = 1 times 3 speed × time = distance △ time = speed 4 unit price × quantity = total price ×

The determinant of n * n matrix is proved to be 0 by mathematical induction

When n = 2, obviously
If n = k, then if n = K + 1, let | a | a | be a determinant of order k + 1 with the same two rows, and only need to prove | a | = 0
In fact, let line I of a be the same as line j, and expand | a | according to the first column_ If the algebraic cofactor of {L1} (L is not equal to I, J) is 0, then | a | = a_ {i1}A_ {i1}+a_ {j1}A_ Since line I of a is the same as line j of a, then a_ {i1}=a_ {j 1}, and a_ {i1}=-A_ If {j 1}, then | a | = 0

Please move a match to make the equation true The following formula is wrong. If you move a match stick in each question, you can make the equation true. How to move it? 1+1+111=2 14+1-1+1=4

1 + 1 + 111 = 2 move 1-1 + 1 + 1 = 2 (move the match with the plus sign in the first 1 + 1 to the middle of 111 to make the plus sign)
14 + 1-1 + 1 = 4 after moving 114 + 1-111 = 4 (move the + sign in 1 + 1 to the front of 14)

Fill in the operation symbol to make the equation true 9 9 9 9 9=16 9 9 9 9 9=17 9 9 9 9 9=18 9 9 9 9 9=19 9 9 9 9 9=20

9 + 9 - (9 + 9) ÷ 9 = 16 idea: 18-2 (two 9's can get 18, then try to use 3 9 to get 2) [(9 + 9) × 9-9] / 9 = 17 idea: 17 = 18-1 = (18 × 9-9) × 9 = 18 idea: using four 9 to get 2 [(9 + 9) × 9 + 9] / 9 = 19 idea: 19 = 18 + 1 = (18 × 9 + 9) △ 9, and calculation

Put 15 matches into 1 + 11 + 111 = 4. Please move one match to make the equation hold

1 + 1 + 1 + 1 = 4 take one of 11 to the tenth of 111

Olympiad Mathematics (moving a matchstick) 1+11+111=12 1+11+111=4 2 is 5 matchsticks, 4 is 4 matchsticks. Move one matchstick to make the equation true

1+11+111=12 --------->>>1+1-1+11=12
1+11+111=4 --------->>>1+1+1+1=4
Move one of the 11th ones to the one in the middle of 111

What is the n power of (ABC) As the title

be equal to
The nth power of a times the nth power of B times the nth power of C

It is known that A.B.C is the length of the three sides of the triangle ABC, and the square of a + 2 B + the square of C = 2B (a + C), try to judge the shape of the triangle That's it. Please tell me

∵ the square of a + the square of 2 B + the square of C = 2B (a + C)
∴a^2+2b^2+c^2－2b(a+c)=0
a^2-2ab+b^2+b^2-2bc+c^2=0
(a-b)^2+(b-c)^2=0
Thus A-B = 0, B-C = 0
We get a = B, B = C
So a = b = C
The shape of the triangle is an equilateral triangle

1. It is known that a, B, C are the three sides of the triangle ABC, and satisfy the relationship a ^ 2 + C ^ 2 = 2Ab + 2bc-2b ^ 2. Try to show that triangle ABC is an equilateral triangle 2. If a, B, C are the three sides of a triangle and satisfy the relation a ^ 2 + B ^ 2 + C ^ 2-ab-ac-bc = 0, try to judge the shape of the triangle

A ^ 2-2ab + B ^ 2 = - C ^ 2-B ^ 2 + 2BC (a-b) ^ 2 = (C-B) ^ 2 (a-b) ^ 2 (a-b) ^ 2 + (B-C) ^ 2 = 0obviously a = b = C triangle ABC is equiltriangle 2. A ^ 2 + B ^ 2 + C ^ 2-ab-ac-bc = 02 * (a ^ 2 + B ^ 2 + C ^ 2-ab-ac-bc = 02 * (a ^ 2 + B ^ 2 + C ^ 2-ab-ac-bc) = 0 (a-b) ^ 2 + (A-C) 2 + (B-C) ^ 2 = 2 = 0 obviously, obviously a = b = C triangle ABC is an equilequilaterals, a = b = C triangle ABC is a equilateral triangle ABC is a equilateral triangle ABC is a three