Ta có:

\(290=2\cdot5\cdot29\\ 895=5\cdot197\\ 578=2\cdot17^2\\ =>ƯCLN\left(290;895;578\right)=1\)

\(290=2.5.29\)

\(895=5.179\)

\(578=2.17^2\)

Nên không tồn tại \(UCLN\left(290;895;578\right)\)

Ta có: `997.1001 `

`= (999 - 2) . (999 + 2) `

`= 999 . 999 - 2 . 999 + 2 . 999 - 4`

`= 999 . 999 - 4 < 999 . 999`

Vậy  `997.1001  < 999 . 999`

\(999.999=\left(1000-1\right)\left(1000-1\right)=\left(1000-1\right)^2=1000^2-2.1000+1\)

\(997.1001=\left(1000-3\right)\cdot\left(1000+1\right)=1000^2-2.1000-2\)

mà \(1>-2\Rightarrow1000^2-2.1000+1>1000^2-2.1000-2\)

\(\Rightarrow999.999>997.1001\)

1

Gọi x là số chia (x ∈ N, x > 15)

Do số dư là 15 nên 224 - 15 = 209 ⋮ x

Ta có:

209 = 11 . 19

Do x > 15 nên x = 19

Vậy số chia là 19, thương là 11

1234567890^2

1.524157875x10¹⁸

\(5^2\cdot x-2^4\cdot x=3^4-6\cdot3^2\\ \Rightarrow25x-16x=81-6\cdot9\\ \Rightarrow9x=81-54\\ \Rightarrow9x=27\\ \Rightarrow x=27:9\\ \Rightarrow x=3\)

\(5^2\cdot x-2^4\cdot x=3^4-6\cdot3^2\)

=>\(25x-16x=81-6\cdot9=81-54=27\)

=>9x=27

=>x=3

`H = 1 - 5 + 5^2 - 5^3 + ... + 5^100`

`5H = 5 - 5^2 + 5^3 - 5^4 + ... + 5^101`

`5H + H = (5 - 5^2 + 5^3 - 5^4 + ... + 5^101) + (1 - 5 + 5^2 - 5^3 + ... + 5^100)`

`6H = 1 + (5 - 5) + (5^2 - 5^2) + ... + (5^100 - 5^100) + 5^101`

`6H = 1 +  5^101`

`H = (1 +  5^101)/6`

`12x + 6/7 : 2 = 9x + 8`

`=> 12x + 3/7 - 9x  =  8`

`=> 3x   =  8 -  3/7`

`=> 3x = 53/7`

`=> x = 53/7 : 3`

`=> x = 53/7 . 1/3`

`=> x=  53/21`

Vậy ..

\(12x+\dfrac{6}{7}\times\dfrac{1}{2}=9x+8\)

\(12x+\dfrac{3}{7}=9x+8\)

\(12x=9x+8-\dfrac{3}{7}\)

\(12x=9x+\dfrac{53}{7}\)

\(\Rightarrow\) \(\dfrac{53}{7}=3x\)

\(\Rightarrow\dfrac{53}{7}\div3=x\)

\(\Rightarrow\dfrac{53}{21}=x\)

Vậy \(x=\dfrac{53}{21}\)

Để \(\dfrac{n}{-3}\) là phân số dương thì n<0

mà n nguyên 

nên \(n\in Z^-\)