Multiplies of 3,4,5 => Multiplies of 60 (3.4.5)

So the numbers satisfy is: 60,120,180,240,300,360,420,480.60,120,180,240,300,360,420,480.So there are 8 positive integers satisfy the question

I don't know this question

Because i don't very good at English

Sorry so much.

 Dao Trong Luan was wrong

Do you have evidence against me?

Quoc Tran Anh Le You can not rely on the number of reviews to accuse me of how many nick

?????? what happen in here??

The answer id NO.

At the begining there are 32 black pieces. We will prove that the number of black pieces left on the board is always even.

In the single move, we change a row or column, assume that row or column has k black pieces and 8 - k while pieces, after change all black to white and while to black, that row or column will has 8-k black and k white. The difference between black pieces after a single move is (8 - k) - k = 8 - 2k. The difference is a even (8 - 2k). Therefore, after every move, the black pieces left is alaways even and cannot be one black piece at any time.

The answer id NO.

At the begining there are 32 black pieces. We will prove that the number of black pieces left on the board is always even.

In the single move, we change a row or column, assume that row or column has k black pieces and 8 - k while pieces, after change all black to white and while to black, that row or column will has 8-k black and k white. The difference between black pieces after a single move is (8 - k) - k = 8 - 2k. The difference is a even (8 - 2k). Therefore, after every move, the black pieces left is alaways even and cannot be one black piece at any time.

My answer is no.

At the begining there are 32 black pieces. We will prove that the number of black pieces left on the board is always even.

In the single move, we change a row or column, assume that row or column has k black pieces and 8 - k while pieces, after change all black to white and while to black, that row or column will has 8-k black and k white. The difference between black pieces after a single move is (8 - k) - k = 8 - 2k. The difference is a even (8 - 2k). Therefore, after every move, the black pieces left is alaways even and cannot be one black piece at any time.

2 oranges

Mỗi người sẽ được 2 quả (  nếu được chia đều cho nhau) 

two oranges

 Let \(S\) be the sum of all the \(mn\) numbers in the table. Note that after an operation, each number stay the same or turns to its negative. Hence there are at most \(2^{mn}\) tables. So \(S\) can only have finitely many possible values. To make the sum of the numbers in each line nonnegative, just look for a line whose numbers have a negative sum. If no such line exists, then we are done. Otherwise, reverse the sign of all the numbers in the line. Then \(S\) increases. Since \(S\) has finitely many possible values, \(S\) can increase finitely many times. So eventually the sum of the numbers in every line must be nonnegative. 

Let SS be the sum of all the mnmn numbers in the table. Note that after an operation, each number stay the same or turns to its negative. Hence there are at most 2mn2mn tables. So SS can only have finitely many possible values. To make the sum of the numbers in each line nonnegative, just look for a line whose numbers have a negative sum. If no such line exists, then we are done. Otherwise, reverse the sign of all the numbers in the line. Then SS increases. Since SS has finitely many possible values, SS can increase finitely many times. So eventually the sum of the numbers in every line must be nonnegative

Let SS be the sum of all the mnmn numbers in the table. Note that after an operation, each number stay the same or turns to its negative. Hence there are at most 2mn2mn tables. So SS can only have finitely many possible values. To make the sum of the numbers in each line nonnegative, just look for a line whose numbers have a negative sum. If no such line exists, then we are done. Otherwise, reverse the sign of all the numbers in the line. Then SS increases. Since SS has finitely many possible values, SS can increase finitely many times. So eventually the sum of the numbers in every line must be nonnegative. 

x + 2/5 = 1/2

x = 1/2 + 2/5

x = 9/10

x = 0,9

x + \(\dfrac{2}{5}\) = \(\dfrac{1}{2}\)

x = \(\dfrac{1}{2}\) + \(\dfrac{2}{5}\)

x = \(\dfrac{9}{10}\)

X+2/5=1/2

X=1/2-2/5

X=1/10

Good luck

X:1/7=14

X=14x1/7

X=2

x : \(\dfrac{1}{7}\) = 14

x = 14 : \(\dfrac{1}{7}\) 

x = 98

Ta có

Vậy A;cho 10

Give A = 2² + 2⁴ + .... + 2¹⁰⁰

4A = 2⁴ + .... + 2¹⁰⁰ + 2¹⁰²

4A - A = (2⁴ + .... + 2¹⁰⁰ + 2¹⁰²) - (2² + 2⁴ + .... + 2¹⁰⁰)

3A = 2¹⁰² - 2²

A = \(\dfrac{2^{102}-4}{3}\)

Ta có : A = 2² + 2⁴ + 2⁶ + ...... + 2¹⁰⁰

=> 2²A = 2⁴ + 2⁶ + ...... + 2¹⁰⁰ + 2¹⁰²

=> 4A - A = 2¹⁰² - 2²

=> 3A = 2¹⁰² - 4

=> A = \(\dfrac{2^{102}-4}{3}\)

very good  

ab=a(b+c)b(b+c)=ab+acb2+bc

a+cb+c=b(a+c)b(b+c)=ab+bcb2+bc

In here , we have :

b2 + bc = b2 + bc

ab = ab

If a>b

=> ac > bc (when a,b ∈N

)

=> ab>a+cb+c

If a<b

=> ac < bc (when a,b ∈N

)

=> ab<a+cb+c

We can see :

If \(\dfrac{a}{b}< 1\)=> a < b

              => ac < ab

              => a. ( b + c ) < b. ( c + a )

              => \(\dfrac{a}{b}< \dfrac{a+c}{b+c}\)

\(\dfrac{a}{b}=\dfrac{a\left(b+c\right)}{b\left(b+c\right)}=\dfrac{ab+ac}{b^2+bc}\)

\(\dfrac{a+c}{b+c}=\dfrac{b\left(a+c\right)}{b\left(b+c\right)}=\dfrac{ab+bc}{b^2+bc}\)

In here , we have :

b² + bc = b² + bc

ab = ab

If a>b

=> ac > bc (when a,b \(\in N\))

=> \(\dfrac{a}{b}>\dfrac{a+c}{b+c}\)

If a<b

=> ac < bc (when a,b \(\in N\))

=> \(\dfrac{a}{b}< \dfrac{a+c}{b+c}\)

Every positive integer has a unique representation in base 2. This is the same as saying that each positive integer can be written uniquely as a sum of different powers of 2. For example, 1 = 2⁰ ,2 = 2¹ ,3 = 2⁰ + 2² , and so on. Recall that we are interested in the numbers beginning x₁ = 3⁰ , x₂ = 3¹ , and x₃ = 3⁰ + 3¹ . We can find the n^th term in this sequence by writing n as a sum of different powers of 2 and then replacing each 2^j in the sum by 3^j . Since n = 100 can be written in base 2 as n = (1100100)₂, we get 100 = 2² + 2⁵ + 2⁶ so that x₁₀₀ = 3² + 3⁵ + 3⁶ = 9 + 243 + 729 = 981

x100 = x98 + x99 = 3^97 + 3^98 = 3.(3^96+3^97)