\(\Rightarrow A=\frac{6n+2-5}{3n+1}=\frac{2\left(3n+1\right)}{3n+1}-\frac{5}{3n+1}\)=\(2-\frac{5}{3n+1}\)

Để A có giá trị nguyên \(\Leftrightarrow5⋮3n+1\Rightarrow3n+1\in\left\{-5;-1;1;5\right\}\)

\(\Rightarrow3n\in\left\{-6;-2;0;4\right\}\Rightarrow n\in\left\{-2;-\frac{2}{3};0;\frac{4}{3}\right\}\) Mà n \(\in Z\) 

\(\Rightarrow n\in\left\{-2;0\right\}\)

Trả lời:

Ta có: \(\frac{6n-3}{3n+1}=\frac{2\left(3n+1\right)-5}{3n+1}=\frac{2\left(3n+1\right)}{3n+1}-\frac{5}{3n+1}=2-\frac{5}{3n+1}\)

 Để A là số nguyên thì \(\frac{5}{3n+1}\)là số nguyên

=> \(5⋮3n+1\) hay \(3n+1\inƯ\left(5\right)\)\(=\left\{\pm1;\pm5\right\}\)

 Ta có bảng sau:

Vậy n \(\in\){ 0 ; -2 } thì A có giá trị nguyên

\(D=200-199-198-...-1\)

\(D=200-\left(199+198+...+2+1\right)\)

\(D=200-\frac{199.200}{2}=-19700\)

a) \(M=2020+2020^2+...+2020^{10}\)

\(M=\left(2020+2020^2\right)+\left(2020^3+2020^4\right)+...+\left(2020^9+2020^{10}\right)\)

\(M=2020\left(1+2020\right)+2020^3\left(1+2020\right)+...+2020^9\left(1+2020\right)\)

\(M=2021\left(2020+2020^3+...+2020^9\right)⋮2021\).

b) Bạn làm tương tự câu a). 

b, \(A=2021+2021^2+...+2021^{2020}\)

\(=2021\left(1+2021\right)+...+2021^{2019}\left(1+2021\right)\)

\(=2022\left(2021+...+2021^{2019}\right)⋮2022\)

Vậy ta có đpcm 

ahihi

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2018}+\frac{1}{2019}\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}+\frac{1}{2019}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}+\frac{1}{2019}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)

\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}\)

=> \(\frac{A}{B}=\frac{\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}}{\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}}=1\)

Trả lời:

\(\frac{x+1}{2}=\frac{x-1}{3}\)

=> 3 ( x + 1 ) = 2 ( x - 1 )

=> 3x + 3 = 2x - 2

=> 3x - 2x = - 2 - 3 

=> x = -6

Vì \(ƯCLN\left(a,b\right)=16\)nên ta đặt \(a=16m,b=16n\)\(\left(m,n\right)=1\).

\(a+b=16m+16n=16\left(m+n\right)=128\Leftrightarrow m+n=8\)

mà \(\left(m,n\right)=1\)nên ta có bảng giá trị: