( x - 2019 )( x - 2020 ) = 0
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VH
1
XO
5 tháng 1 2020
(x-2020)x - 1 - (x - 2020)x + 2019 = 0
=> (x - 2020)x - 1 .[(x - 2020)2020 - 1] = 0
=> \(\orbr{\begin{cases}\left(x-2020\right)^{x-1}=0\\\left(x-2020\right)^{2020}-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x-2020=0\\\left(x-2020\right)^{2020}=1^{2020}\end{cases}\Rightarrow}\orbr{\begin{cases}x-2020=0\\x-2020=\pm1\end{cases}}}\)
=> \(x-2020\in\left\{0;1;-1\right\}\Rightarrow x\in\left\{2020;2021;2019\right\}\)
TY
2
DH
4 tháng 12 2019
Ta có: \(\left|x-\frac{2018}{2019}\right|\ge0\)
Và: \(\left|x-\frac{2019}{2020}\right|\ge0\)
\(\Rightarrow\left|x-\frac{2018}{2019}\right|+\left|x-\frac{2019}{2020}\right|\ge0\)
\(\Rightarrow\left|x-\frac{2018}{2019}\right|+\left|x-\frac{2019}{2020}\right|=0\)
\(\Leftrightarrow\left|x-\frac{2018}{2019}\right|=\left|x-\frac{2019}{2020}\right|=0\left(Vônghiệm\right)\)
VM
5 tháng 12 2019
\(\left|x-\frac{2018}{2019}\right|+\left|x-\frac{2019}{2020}\right|=0\)
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!
21 tháng 10 2021
\(x^2+x-2019\cdot2020=0\)
\(\Leftrightarrow\left(x+2020\right)\left(x-2019\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2020\\x=2019\end{matrix}\right.\)
9 tháng 5 2021
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 ...
HH
1
23 tháng 6 2019
\(\left(x+2018\right)^{2020}-\left(x+2018\right)^{2019}=0\)
\(\Leftrightarrow\) \(\left(x+2018\right)^{2019}\left(x+2018-1\right)=0\)
\(\Leftrightarrow\) \(\left(x+2018\right)^{2019}\left(x+2017\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+2018\right)^{2019}=0\\x+2017=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2018\\x=-2017\end{matrix}\right.\)
NT
23 tháng 6 2019
\(\left(x+2018\right)^{2020}-\left(x+2018\right)^{2019}=0\\ \Leftrightarrow\left(x+2018\right)^{2019}\left[\left(x+2018\right)^2-1\right]=0\\ \Leftrightarrow\left(x+2018\right)^{2019}\left(x+2017\right)\left(x+2019\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+2018=0\\x+2017=0\\x+2019=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2018\\x=-2017\\x=-2019\end{matrix}\right.\)
`( x - 2019 )( x - 2020 ) = 0`
`@ TH1`
`x-2019=0`
`x=0+2019`
`x=2019`
` @ TH2`
`x-2020=0`
`x=0+2020`
`x=2020`
\(\text{( x – 2019 ) ( x – 2020 ) = 0}\)
\(\text{+) x – 2019 = 0 }\)\(\Rightarrow\text{x = 2019}\)
\(\text{+) x – 2020 = 0}\) \(\Rightarrow\text{x = 2020}\)