ĐKXĐ : \(\left\{{}\begin{matrix}x>2019\\y>2020\\z>2021\end{matrix}\right.\)

Đặt \(\sqrt{x-2019}=a,......\)

Ta được PT : \(\dfrac{1-a}{a^2}+\dfrac{1-b}{b^2}+\dfrac{1-c}{c^2}+\dfrac{3}{4}=0\)

\(\Leftrightarrow\dfrac{1}{a^2}-\dfrac{1}{a}+\dfrac{1}{4}+\dfrac{1}{b^2}-\dfrac{1}{b}+\dfrac{1}{4}+\dfrac{1}{c^2}-\dfrac{1}{c}+\dfrac{1}{4}=0\)

\(\Leftrightarrow\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{b}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{c}-\dfrac{1}{2}\right)^2=0\)

- Thấy : \(\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2\ge0,......\)

\(\Rightarrow\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{b}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{c}-\dfrac{1}{2}\right)^2\ge0\)

- Dấu " = " xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{2}\\\dfrac{1}{b}=\dfrac{1}{2}\\\dfrac{1}{c}=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)

- Thay lại a. b. c ta được : \(\left\{{}\begin{matrix}\sqrt{x-2019}=2\\\sqrt{y-2020}=2\\\sqrt{z-2021}=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2019=4\\y-2020=4\\z-2021=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2023\\y=2024\\z=2025\end{matrix}\right.\) ( TM )

Vậy ...

=>\(\left(\dfrac{x+1}{2021}+1\right)+\left(\dfrac{x+2}{2020}+1\right)+\left(\dfrac{x+3}{2019}+1\right)+\left(\dfrac{x+2028}{2}-3\right)=0\)

=>x+2022=0

=>x=-2022

\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+....+\dfrac{1}{\left(x+2020\right)\left(x+2021\right)=1}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+...+\dfrac{1}{x+2020}-\dfrac{1}{x+2021}=1\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2021}=1\)

\(\Leftrightarrow\dfrac{\left(x+2021\right)-\left(x+1\right)}{\left(x+1\right)\left(x+2021\right)}=1\)

\(\Leftrightarrow\dfrac{x+2021-x-1}{\left(x+1\right)\left(x+2021\right)}=1\)

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

\(\Leftrightarrow\left(x+1\right)\left(x+2021\right)=2020\)

\(\Leftrightarrow x^2+2021x+x+2021=2020\)

\(\Leftrightarrow x^2+2022x=-1\)

\(\Leftrightarrow x\left(x+2022\right)=-1\)

Đến đây bạn chia trường hợp để giaỉ ra nghiệm nguyên nhé

\(\dfrac{x-1}{2023}+\dfrac{x-2}{2022}=\dfrac{x-3}{2021}+\dfrac{x-4}{2020}\)

`<=>(x-1)/2023-1+(x-2)/2022-1=(x-3)/2021-1+(x-4)/2020-1`

`<=>(x-2024)/2023+(x-2024)/2022=(x-2024)/2021+(x-2024)/2020`

`<=>(x-2024)(1/2023+1/2022-1/2021-1/2020)=0`

`<=>x-2024=0(1/2023+1/2022-1/2021-1/2020>0)`

`<=>x=2024`

=>\(\left(\dfrac{x-1}{2023}-1\right)+\left(\dfrac{x-2}{2022}-1\right)=\left(\dfrac{x-3}{2021}-1\right)+\left(\dfrac{x-4}{2020}-1\right)\)

=>x-2024=0

=>x=2024

\(\dfrac{x-1}{2023}+\dfrac{x-2}{2022}=\dfrac{x-3}{2021}+\dfrac{x-4}{2020}\)

⇔\(\dfrac{x-1}{2023}-1+\dfrac{x-2}{2022}-1=\dfrac{x-3}{2021}-1+\dfrac{x-4}{2020}\)

⇔\(\dfrac{x-1}{2023}-\dfrac{2023}{2023}+\dfrac{x-2}{2022}-\dfrac{2022}{2022}=\dfrac{x-3}{2021}-\dfrac{2021}{2021}+\dfrac{x-4}{2020}-\dfrac{2020}{2020}\)

⇔\(\dfrac{x-2024}{2023}+\dfrac{x-2024}{2022}=\dfrac{x-2024}{2021}+\dfrac{x-2024}{2020}\)

⇔\(\dfrac{x-2024}{2023}+\dfrac{x-2024}{2022}-\dfrac{x-2024}{2021}-\dfrac{x-2024}{2020}=0\)

⇔\(\left(x-2024\right)\left(\dfrac{1}{2023}+\dfrac{1}{2022}-\dfrac{1}{2021}-\dfrac{1}{2020}\ne0\right)\)

⇔\(x-2024=0\)

⇔\(x=2024\)

xét x=y,x>y và x<y chú ý tới điều kiện x,y thuộc -1;1 nữa 

a) \(3\left(2x-x\right)=5x+1\)

\(\Leftrightarrow6x-3x=5x+1\)

\(\Leftrightarrow6x-3x-5x=1\)

\(\Leftrightarrow-2x=1\)

\(\Leftrightarrow x=\dfrac{1}{-2}=-\dfrac{1}{2}\)

b) \(\dfrac{x+1}{2021}+\dfrac{x+2}{2020}+\dfrac{x+3}{2019}+\dfrac{x+4}{2018}=0\)

\(\Leftrightarrow\dfrac{x+1}{2021}+1+\dfrac{x+2}{2020}+1=\dfrac{x+3}{2019}+1+\dfrac{x+4}{2018}+1\)

\(\Leftrightarrow\dfrac{x+2022}{2021}+\dfrac{x+2022}{2020}=\dfrac{x+2022}{2019}+\dfrac{x+2022}{2018}\)

\(\Leftrightarrow\left(x+2022\right)\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}\right)\)

\(\Leftrightarrow x+2022=0\)

\(\Leftrightarrow x=-2022\)

a)3(2x-3)=5x+1

⇔6x-9=5x+1

⇔6x-5x=1+9

⇔x=10

vậy phương trình có nghiệm là S={10}

b)\(\dfrac{x+1}{2021}\)+\(\dfrac{x+2}{2020}\)+\(\dfrac{x+3}{2019}\)+\(\dfrac{x+2028}{2}\)=0

⇔2020(x+1)+2021(x+2)+2041210(x+2028)=0

⇔2045251x+4139579942=0

⇔2045251x=-4139579942=0

⇔x=-\(\dfrac{4139579942}{2045251}\)

vậy phương trình có tập nghiệm là S={\(-\dfrac{4139579942}{2045251}\)}