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(x-2020)^{x - 1} - (x - 2020)^{x + 2019} = 0

=> (x - 2020)^{x - 1} .[(x - 2020)²⁰²⁰ - 1] = 0 

=> \(\orbr{\begin{cases}\left(x-2020\right)^{x-1}=0\\\left(x-2020\right)^{2020}-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x-2020=0\\\left(x-2020\right)^{2020}=1^{2020}\end{cases}\Rightarrow}\orbr{\begin{cases}x-2020=0\\x-2020=\pm1\end{cases}}}\)

=> \(x-2020\in\left\{0;1;-1\right\}\Rightarrow x\in\left\{2020;2021;2019\right\}\)

Ta có: \(\left|x-\frac{2018}{2019}\right|\ge0\)

Và: \(\left|x-\frac{2019}{2020}\right|\ge0\)

\(\Rightarrow\left|x-\frac{2018}{2019}\right|+\left|x-\frac{2019}{2020}\right|\ge0\)

\(\Rightarrow\left|x-\frac{2018}{2019}\right|+\left|x-\frac{2019}{2020}\right|=0\)

\(\Leftrightarrow\left|x-\frac{2018}{2019}\right|=\left|x-\frac{2019}{2020}\right|=0\left(Vônghiệm\right)\)

\(\left|x-\frac{2018}{2019}\right|+\left|x-\frac{2019}{2020}\right|=0\)

Ta có:

\(\left\{{}\begin{matrix}\left|x-\frac{2018}{2019}\right|\ge0\\\left|x-\frac{2019}{2020}\right|\ge0\end{matrix}\right.\forall x.\)

\(\Rightarrow\left|x-\frac{2018}{2019}\right|+\left|x-\frac{2019}{2020}\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x-\frac{2018}{2019}\right|=0\\\left|x-\frac{2019}{2020}\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-\frac{2018}{2019}=0\\x-\frac{2019}{2020}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{2018}{2019}\\x=\frac{2019}{2020}\end{matrix}\right.\)

\(\Rightarrow\) Vô lí vì x không thể đồng thời nhận 2 giá trị khác nhau.

Vậy không tồn tại giá trị nào của x thỏa mãn yêu cầu đề bài.

Chúc bạn học tốt!

\(x^2+x-2019\cdot2020=0\)

\(\Leftrightarrow\left(x+2020\right)\left(x-2019\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2020\\x=2019\end{matrix}\right.\)

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 ...

Đề bài yêu cầu gì?

<=> (x² +2019x)-(2020x+2019.2020)=0
<=> x.(x+2019)-2020.(x+2019)=0
<=>(x-2020).(x+2019)=0
câu kia tương tự

\(\left(x+2018\right)^{2020}-\left(x+2018\right)^{2019}=0\)

\(\Leftrightarrow\) \(\left(x+2018\right)^{2019}\left(x+2018-1\right)=0\)

\(\Leftrightarrow\) \(\left(x+2018\right)^{2019}\left(x+2017\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+2018\right)^{2019}=0\\x+2017=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2018\\x=-2017\end{matrix}\right.\)

\(\left(x+2018\right)^{2020}-\left(x+2018\right)^{2019}=0\\ \Leftrightarrow\left(x+2018\right)^{2019}\left[\left(x+2018\right)^2-1\right]=0\\ \Leftrightarrow\left(x+2018\right)^{2019}\left(x+2017\right)\left(x+2019\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+2018=0\\x+2017=0\\x+2019=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2018\\x=-2017\\x=-2019\end{matrix}\right.\)