Đk: \(\forall x\in R\)

Ta có:\(\sqrt{x^2+1-2x}+\sqrt{x^2+4x+4}=\sqrt{1+2020^2+\frac{2020^2}{2021^2}}+\frac{2020}{2021}\)

<=> \(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}=\sqrt{1+2020^2+2.2020+\frac{2020^2}{2021^2}-2.2020}+\frac{2020}{2021}\)

<=> \(\left|x-1\right|+\left|x+2\right|=\sqrt{\left(1+2020\right)^2+\frac{2020^2}{2021^2}-2.2020}+\frac{2020}{2021}\)

<=> \(\left|x-1\right|+\left|x+2\right|=\sqrt{\left(2021-\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)

<=> \(\left|x-1\right|+\left|x+2\right|=\frac{2021^2-2020}{2021}+\frac{2020}{2021}\)

<=> \(\left|x-1\right|+\left|x+2\right|=2021\)

Lập bảng xét dầu

x                   -2                   1 

x - 1   -         |           -          0       +

x + 2   -        0         +          |            -

Xét các TH xảy ra :

TH1: x \(\le\)-2 => pt trở thành: 1 - x - x - 2 = 2021

<=> -2x = 2022 <=> x = -1011 (tm)

TH2: \(-2< x\le1\) => pt trở thành: 1 - x + x + 2 = 2021

<=> 0x = 2018 (vô lí) => pt vô nghiệm

TH3: \(x>1\) => pt trở thành: x - 1 + x + 2 = 2021

<=> 2x = 2020 <=> x = 1010 (tm)

Vậy S = {-1011; 1010}

ko ghi lại đề 

ta thấy : 2019 - 1 = 2018 

2020 - 2 = 2018 

2021 - 3 = 2018 

2022 - 4 = 2018 

=> x = 2018

thử lại :

2018+1/2019 + 2018+2/2020 = 2018+3/2021 + 2018+4/2022

= 1 + 1 = 1 + 1

2 = 2

2020 - 2 = 2018 
2021 - 3 = 2018 
2022 - 4 = 2018 
=> x = 2018

thây zô mà thử lại

\(\frac{x+4}{2019}+\frac{x+3}{2020}=\frac{x+2}{2021}+\frac{x+1}{2020}\)

\(\Leftrightarrow(\frac{x+4}{2019}+1)+(\frac{x+3}{2020}+1)=(\frac{x+2}{2021}+1)+(\frac{x+1}{2022}+1)\)

\(\Leftrightarrow\frac{x+2023}{2019}+\frac{x+2023}{2020}=\frac{x+2023}{2021}+\frac{x+2023}{2022}\)

\(\Leftrightarrow\frac{x+2023}{2019}+\frac{x+2023}{2020}-\frac{x+2023}{2021}-\frac{x+2023}{2022}=0\)

\(\Leftrightarrow\left(x+2023\right)\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}-\frac{1}{2020}\right)=0\)

\(\Leftrightarrow x+2023=0\)

\(\Leftrightarrow x=-2023\)

Nhầm đề :( Với bước thứ 4 sửa thành ( 1/2019 + 1/2020 - 1/2021 - 1/2022 ) 

\(\frac{x+4}{2019}+\frac{x+3}{2020}=\frac{x+2}{2021}+\frac{x+1}{2022}\)

\(\Leftrightarrow\)\(\frac{x+4}{2019}+\frac{x+3}{2020}+2=\frac{x+2}{2021}+\frac{x+1}{2022}+2\)

\(\Leftrightarrow\)\(\left(\frac{x+4}{2019}+1\right)+\left(\frac{x+3}{2020}+1\right)=\left(\frac{x+2}{2021}+1\right)+\left(\frac{x+1}{2022}+1\right)\)

\(\Leftrightarrow\)\(\left(\frac{x+4}{2019}+\frac{2019}{2019}\right)+\left(\frac{x+3}{2020}+\frac{2020}{2020}\right)\)\(=\)\(\left(\frac{x+2}{2021}+\frac{2021}{2021}\right)+\left(\frac{x+1}{2022}+\frac{2022}{2022}\right)\)

\(\Leftrightarrow\)\(\frac{x+2023}{2019}+\frac{x+2023}{2020}=\frac{x+2023}{2021}+\frac{2023}{2022}\)

\(\Leftrightarrow\)\(\frac{x+2023}{2019}+\frac{x+2023}{2020}-\frac{x+2023}{2021}-\frac{x+2023}{2022}=0\)

\(\Leftrightarrow\) \(\left(x+2023\right).\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}-\frac{1}{2022}\right)=0\)

\(\Leftrightarrow\)\(x+2023=0\)      ( Vì \(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}-\frac{1}{2022}\ne0\))

\(\Leftrightarrow\)\(x=-2023\)

Vậy x = -2023

Câu 1 :

a) ĐKXĐ : \(\hept{\begin{cases}x+1\ne0\\2x-6\ne0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)

b) Để \(P=1\Leftrightarrow\frac{4x^2+4x}{\left(x+1\right)\left(2x-6\right)}=1\)

\(\Leftrightarrow\frac{4x^2+4x-\left(x+1\right)\left(2x-6\right)}{\left(x+1\right)\left(2x-6\right)}=0\)

\(\Rightarrow4x^2+4x-2x^2+4x+6=0\)

\(\Leftrightarrow2x^2+8x+6=0\)

\(\Leftrightarrow x^2+4x+4-1=0\)

\(\Leftrightarrow\left(x+2-1\right)\left(x+2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=-3\left(TMĐKXĐ\right)\end{cases}}\)

Vậy : \(x=-3\) thì P = 1.