Khoảng cách : `1`

Số hạng:

`(50-1)/1 + 1 = 50(số-hạng)`

Tổng:

`(50+1)xx50:2=1275`

Vậy tổng `A` là `1275`

`#LeMichael`

Số số hạng của dãy A là: 

\(\left(50-1\right):1+1=50\)(số)

\(\Rightarrow A=\dfrac{\left(50+1\right)50}{2}=1275\)

Vậy \(A=1275\)

Ta có:

\(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\)

\(\Leftrightarrow\dfrac{x+1}{65}+1+\dfrac{x+3}{63}+1=\dfrac{x+5}{61}+1+\dfrac{x+7}{59}+1\)

\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}=\dfrac{x+66}{61}+\dfrac{x+66}{59}\)

\(\Leftrightarrow\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\)

\(\Leftrightarrow x+66=0\Leftrightarrow x=-66\)

27 = 3.3.3

21= 7.3

Chắc chắn b là số nguyên dương có

a = b.21 + 7.3 = 3 ( 9b + 7)

a chia hết cho 3 nhưng không chia hết cho 9

\(\dfrac{x-3}{13}+\dfrac{x-3}{14}=\dfrac{x-3}{15}+\dfrac{x-3}{16}\)

\(\Leftrightarrow\dfrac{x-3}{13}+\dfrac{x-3}{14}-\dfrac{x-3}{15}-\dfrac{x-3}{16}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)

\(\Leftrightarrow x-3=0\) (vì\(\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{15}-\dfrac{1}{16}>0\) (*) )

\(\Leftrightarrow x=3\)

Chứng minh (*):

Vì\(13< 15\Rightarrow\dfrac{1}{13}>\dfrac{1}{15}\Rightarrow\dfrac{1}{13}-\dfrac{1}{15}>0\)

\(14< 16\Rightarrow\dfrac{1}{14}>\dfrac{1}{16}\Rightarrow\dfrac{1}{14}-\dfrac{1}{16}>0\)

Do đó:\(\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{15}-\dfrac{1}{16}>0\)

=> (*) luôn đúng

Vậy x = 3 là giá trị cần tìm.

x-3/13+x-3/14=x-3/15+x-3/16

<=> x-3/13+x-3/14-x-3/15-x-3/16=0

<=> (x-3). (1/13+1/14-1/15-1/16)

<=> (x-3)=0 ( Vì 1/13+1/14-1/15-1/16>0)

<=> x-3=0 => x=3

Vậy x=3

`x+2.x+3.x+4.x=3708.100`

`1.x+2.x+3.x+4.x=370800`

`(1+2+3+4).x = 370800`

`10.x = 370800`

`x=370800:10`

`x=37080`

\(\dfrac{17}{103}< \dfrac{19}{113}\)

\(3+3^2+3^3+3^4\)

\(=3\left(1+3+3^2+3^3\right)\)

\(=3\cdot40\)

\(=3\cdot10\cdot4⋮4\left(dpcm\right)\)

\(2+2^2+2^3+....+2^{2022}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2021}+2^{2022}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2021}\left(1+2\right)\)

\(=2\cdot3+2^3\cdot3+...+2^{2021}\cdot3\)

\(=3\left(2+2^3+...+2^{2011}\right)⋮3\left(dpcm\right)\)

\(S=2+2^2+2^3+...+2^{10}\) 

\(S=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\) (5 cặp )

\(S=2\left(1+2\right)+2^3\left(1+2\right)+..+2^9\left(1+2\right)\)

\(S=2\cdot3+2^3\cdot3+...+2^9\cdot3\)

\(=3\left(2+2^3+..+2^9\right)⋮3\left(dpcm\right)\)