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`Answer:`

\(\frac{x+721}{2020}+\frac{x+21}{700}+\frac{x+721}{2021}=-1\) (Mình bổ sung đề nhé.)

\(\Leftrightarrow\frac{x+721}{2020}+[\frac{x+21}{700}+1]+\frac{x+721}{2021}=0\)

\(\Leftrightarrow\frac{x+721}{2020}+\frac{x+721}{700}+\frac{x+721}{2021}=0\)

\(\Leftrightarrow\left(x+721\right).\left(\frac{1}{2020}+\frac{1}{700}+\frac{1}{2021}\right)=0\)

Mà \(\frac{1}{2020}+\frac{1}{700}+\frac{1}{2021}\ne0\)

\(\Rightarrow x+721=0\Leftrightarrow x=-721\)

`  (x+721) /2020+(x+21) /700+ (x+721)/2021=-1`

`<=>(x+721)/2020+(x+21) /700+ 1+ (x+721)/2021=0`

`<=>(x+721)/2020+(x+721) /700+ (x+721)/2021=0 `

`<=>(x +721) (1/2020+1/700+1/2021 )= 0` 

Vì `1/2020+1/700+1/2021>0` 

 `=>x+721 =0 <=>x=-721`

Vậy `S={-721}` 

Ta có: \(\dfrac{x+721}{2020}+\dfrac{x+21}{700}+\dfrac{x+721}{2021}=-1\)

\(\Leftrightarrow\dfrac{x+721}{2020}+\dfrac{x+721}{700}+\dfrac{x+721}{2021}=0\)

\(\Leftrightarrow\left(x+721\right)\left(\dfrac{1}{2020}+\dfrac{1}{700}+\dfrac{1}{2021}\right)=0\)

mà \(\dfrac{1}{2020}+\dfrac{1}{700}+\dfrac{1}{2021}>0\)

nên x+721=0

hay x=-721

Vậy: S={-721}

-721

a: \(\dfrac{-11}{18}+\dfrac{12}{29}+\dfrac{-7}{18}+\dfrac{2020}{2021}+\dfrac{17}{29}\)

=(-11/18-7/18)+(12/29+17/29)+2020/2021

=2020/2021

b: \(\dfrac{-2}{3}\cdot\dfrac{4}{9}+\dfrac{5}{9}\cdot\dfrac{-2}{3}+\dfrac{4}{3}\)

=-2/3+4/3=2/3

Số số hạng là: \(\dfrac{770-7}{7}+1=110-1+1=110\left(số\right)\)

Tổng của dãy số là:

\(\left(770+7\right)\cdot\dfrac{110}{2}=42735\)

(x + 721) x 6 = 1286 x 6

(x + 721) x 6 = 7 716

x + 721 = 7 716 : 6

x + 721 = 1286

x = 1286 - 721

x= 565

`@` `\text {Ans}`

`\downarrow`

`x + 257 = 981`

`=> x = 981 - 257`

`=> x = 724`

Vậy, `x = 724`

`x - 546 = 35`

`=> x = 35 + 546`

`=> x = 581`

Vậy, `x = 581`

`721 - x = 615`
`=> x = 721 - 615`

`=> x = 106`

Vậy, `x = 106`

`@` `\text {Kaizuu lv uuu}`

Áp dụng BĐT trị tuyệt đối:

\(M=\left|x-2019\right|+\left|2021-x\right|+2020\left|x-2020\right|\)

\(M\ge\left|x-2019+2021-x\right|+2020\left|x-2020\right|=2+2020\left|x-2020\right|\ge2\)

\(\Rightarrow M_{min}=2\) khi \(\left\{{}\begin{matrix}\left(x-2019\right)\left(2021-x\right)\ge0\\\left|x-2020\right|=0\end{matrix}\right.\) \(\Rightarrow x=2020\)