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

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

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

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!

Ta có x+1/2018 + x+1/2019 + x+1/2020 = 0

=> (x+1).(1/2018 + 1/2019+ 1/2020) = 0

Vì 1/2018 + 1/2019 + 1/2020 > 0

=> x+1 = 0

=> x = -1

Đề bài yêu cầu gì?

a)  (x+3)(x+5)=0

=>x+3=0 hoặc x+5=0

=>x=-3 hoặc -5

b) (x-1).5-1=0

=>5x-5-1=0

=>5x-6=0

=>5x=6

=>x=6/5

c) 

làm câu c,d,b và E đi bạn

OK bạn

Nhận xét : ( x + y - 3 )^2018 >=0 và 2018.(2x-4)^2020 >= 0

=> (x+y-3)^2018 + 2018.(2x-4)^2020 >=0 

Dấu = xảy ra khi : x + y - 3 = 0 và 2x - 4 = 0 => x = 2 và y = 1

Thay vào bt S :

S = ( 2 - 1)^2019 + (2-1)^2019

= 1^2019 + 1^2019 = 2

em cảm ơn

\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}+\dfrac{x+3}{2018}+\dfrac{x+4}{2017}+4=0\)

⇔ \(\dfrac{x+1}{2020}+1+\dfrac{x+2}{2019}+1+\dfrac{x+3}{2018}+1+\dfrac{x+4}{2017}+1=0\)

\(\Leftrightarrow\) \(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}+\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}=0\)

⇔ \(\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+\dfrac{1}{2017}\right)=0\)

\(Do\) \(\left(\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+\dfrac{1}{2017}\right)\ne0\)

⇒ \(x+2021=0\)

⇔ \(x=-2021\)

\(Vậy\) \(x=-2021\)

\(\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2019}x2018\)
\(=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{2018}{2019}=2-\dfrac{2019}{2018}=\dfrac{2017}{2018}\)

D

D

Ta có |x| \(\ge\) 0 \(\forall\) x

\(\Rightarrow\left|x\right|+2020\ge2020\)

D