\(\dfrac{1}{3}\cdot\dfrac{2}{3}+\dfrac{1}{3}\cdot\dfrac{3}{4}-\dfrac{1}{3}\cdot\dfrac{17}{12}\)

\(=\dfrac{1}{3}\left(\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{17}{12}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{8}{12}+\dfrac{9}{12}-\dfrac{17}{12}\right)\)

\(=\dfrac{1}{3}\cdot0=0\)

\(\dfrac{1}{3}.\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{3}{4}-\dfrac{1}{3}.\dfrac{17}{12}\)

= \(\dfrac{1}{3}.\left(\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{17}{12}\right)\)

= \(\dfrac{1}{3}.\left(\dfrac{8}{12}+\dfrac{9}{12}-\dfrac{17}{12}\right)\)

= \(\dfrac{1}{3}.\left(\dfrac{17}{12}-\dfrac{17}{12}\right)\)

= \(\dfrac{1}{3}.0\)

= `0`

\(x^{50}=x\)

=>\(x^{50}-x=0\)

=>\(x\left(x^{49}-1\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(x^{50}=x\)

\(x^{50}-x=0\\\)

\(x\cdot x^{49}-x\cdot1=0\)

\(x\cdot\left(x^{49}-1\right)=0\)

\(x=0\) hoặc \(x^{49}-1=0\)

\(x=0\) hoặc \(x^{49}=1\)

\(x=0\) hoặc \(x=1\)

vậy \(x=0\) hoặc \(x=1\)

Đặt \(A=3^4+3^6+3^8+...+3^{16}+3^{18}\)

\(\Rightarrow9A=3^6+3^8+3^{10}+...+3^{18}+3^{20}\\ \Rightarrow9A-A=\left(3^6+3^8+3^{10}+...+3^{18}+3^{20}\right)-\left(3^4+3^6+3^8+...+3^{16}+3^{18}\right)\\ \Rightarrow8A=3^{20}-3^4\\ \Rightarrow A=\dfrac{3^{20}-3^4}{8}\)

Vậy \(3^4+3^6+3^8+...+3^{16}+3^{18}=\dfrac{3^{20}-3^4}{8}\)

$A=3+3^2+3^3+\dots+3^{100}$

$3A=3^2+3^3+3^4+\dots+3^{101}$

$3A-A=(3^2+3^3+3^4+\dots+3^{101})-(3+3^2+3^3+\dots+3^{100})$

$2A=3^{101}-3$

$\Rightarrow 2A+3=3^{101}$

Mặt khác: $2A+3=3^n$. Do đó: $3^n=3^{101}\Rightarrow n=101$ (tmdk)

\(A=3+3^2+3^3+...+3^{100}\\ 3A=3^2+3^3+...+3^{101}\\ 3A-A=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^2+...+3^{100}\right)\\ 2A=3^{101}-3\\ A=\dfrac{3^{101}-3}{2}\)

Mà: 

\(2A+3=3^n\\ =>2\cdot\dfrac{3^{101}-3}{2}+3=3^n\\ =>3^{101}-3+3=3^n\\ =>3^n=3^{101}\\ =>n=101\)

g;  (\(x-4\))(y + 1) =8

    Ư(8) = {- 8; - 4; - 2; -1; 1; 2; 4; 8}

    Lập bảng ta có: 

Theo bảng trên ta có:

(\(x\); y) = (- 4; - 2); (0; -3); (2; - 5); (3; - 9); (5; 7); (6; 3); (8; 1); (12; 0)

h; (2\(x\) + 3)(y - 2) = 15

   Ư(15) = {- 15; - 5; - 3; - 1; 1; 3; 5; 15}

  Lập bảng ta có:

Theo bảng trên ta có: 

(\(x;y\)) = (- 9; 1); (- 4; - 1); (- 2; - 13); (- 1; 17); (0; 7); (1; 5); (6; 3)

b: Vì 2n+1;2n+2;2n+3 là ba số tự nhiên liên tiếp

nên \(\left(2n+1\right)\left(2n+2\right)\left(2n+3\right)⋮3\)

Từ 2 đến 201 số lượng số hạng là: (201 - 2) : 1 + 1 = 200 (số hạng) 

Số lượng cặp là: 200 : 2 = 100 (cặp)

1 - 2 + 3 - 4 + 5 - ... + 199 - 200 + 201 

= 1 + (-2 + 3) + (-4 + 5) + ... + (-198 + 199) + (-200 + 201) 

= 1 + 1 + 1 + ... + 1 + 1

= 1 + 100*1 

= 1 + 100

= 101 

\(D=4-4^2+4^3-4^4+...+4^{2024}\\ 4D=4^2-4^3+4^4-4^5+...+4^{2025}\\ 4D+D=\left(4^2-4^3+4^4-4^5+...+4^{2025}\right)+\left(4-4^2+4^3-4^4+...+4^{2024}\right)\\ 5D=4^{2025}+4\\ D=\dfrac{4^{2025}+4}{5}\)

Đặt:

\(A=1+3+3^2+...+3^{2023}\\ 3A=3+3^2+3^3+...+3^{2024}\\ 3A-A=\left(3+3^2+3^3+..+3^{2024}\right)-\left(1+3+3^2+...+3^{2023}\right)\\ 2A=3^{2024}-1\\ A=\dfrac{3^{2024}-1}{2}\)

`A =` \(1+3+3^2+...+3^{2023}\)

\(3A=3+3^2+3^3+...+3^{2024}\)

`3A - A =` \(\left(3+3^2+3^3+...+3^{2024}\right)-\left(1+3+3^2+...+3^{2023}\right)\)

`2A =` \(3^{2024}-1\)

`A =` \(\dfrac{3^{2024}-1}{2}\)