Với x < 2017 

pt <=> (2017 - x) + 2018 - x + 2019 - x = 2

    <=> 6054 - 3x = 2

    <=> 3x = 6054 - 2 = 6052

    <=>  x = \(\frac{6052}{3}>2017\) (Loại)

Với \(2017\le x\le2018\)

pt <=> (x - 2017) + (2018 - x) + (2019 - x) = 2

    <=>  2020 - x = 2

    <=>  x = 2020 - 2 = 2018 (Nhận) 

Với \(2018< x\le2019\)

pt <=> (x - 2017) + (x - 2018) + (2019 - x) = 2 

    <=>  x - 2016 = 2

    <=>  x = 2018  (loại)

Với \(2019< x\)

pt <=> (x - 2017) + (x - 2018) + (x - 2019) = 2 

    <=> 3x - 6054 = 2

    <=>  3x = 6056

    <=> x = \(\frac{6056}{3}< 2019\) (Loại )

Vậy , phương trình chỉ có một nghiệm x = 2018 

|2017-x|+|2018-x|+|2019-x|=2

nên sẽ có ít nhất 1 giá trị bằng 0

1. |2017-x|=0

2017-x=0

x=2017

=>|2017-x|+|2018-x|+|2019-x|=3(không thỏa mãn)

2.|2018-x|=0

2018-x=0

x=2018

=>|2017-x|+|2018-x|+|2019-x|=2(thỏa mãn)

3.|2019-x|=0

2019-x=0

x=2019 =>|2017-x|+|2018-x|+|2019-x|=3(không thỏa mãn) Vậy x=2018 để thỏa mãn điều kiện|2017-x|+|2018-x|+|2019-x|=2

꧁༺ⓂⓉⓅ_ⓀⒶⒾⓉⓄ༻꧂( ༺TEAM༻❺❾☆ⓇⓄⓎⒶⓁ )

chép mạng nhớ ghi nguồn nha 

https://h7.net/hoi-dap/toan-7/tim-x-biet-2017-x-2018-x-2019-x-2-faq358792.html

\(\dfrac{x-1}{2019}+\dfrac{x-2}{2018}+\dfrac{x-3}{2017}=3\)

\(\Leftrightarrow\left(\dfrac{x-1}{2019}-1\right)+\left(\dfrac{x-2}{2018}-1\right)+\left(\dfrac{x-3}{2017}-1\right)=0\)

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

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

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

Vi \(\dfrac{1}{2019}+\dfrac{1}{2018}+\dfrac{1}{2017}\ne0\)

nên \(x-2020=0\)

\(\Leftrightarrow x=2020\)

Vậy ...

=>|x-2017|+|2018-x|+|2019-x|=2(mỗi s/h < =2)                           TH1;|2019-x|=0=>2019-x=0                                                           

 ta có; |x-2017|+|2018-x|+|2019-x| >= |x-2017+2018-x|+|2019-x|                        =>x=2019=>tích =3(L)

=>                                      >= |1|+|2019-x|=1+|2019-x|            TH2;|2019-x|=1=>hoặc2019-x=1;hoặc = -1                                          => 2                                     >= 1+|2019-x|                                                 =>hoặc x=2018;hoặc = 2020

 => 1                                         >= |2019-x|                                                 =>hoặc tích=2(TM);tích=6(L)                                                                                                                                      Vậy x=2018

=>|2019-x|={1;0}

viết nhầm ; "tích" sửa thành "tổng"

ta có |2017-x|+|2019-x|=|2017-x|+|x-2019|>=|2017-x+x-2019|=|-2|=2

=>|2017-x|+|x-2019|>=2

Dấu "=" xảy ra khi (2017-x)(x-2019)>=0

<=>\(\orbr{\begin{cases}\hept{\begin{cases}2017-x\le0\\x-2019\le0\end{cases}}\\\hept{\begin{cases}2017-x>0\\x-2019>0\end{cases}}\end{cases}}\)

thui mỏi tay quá,tự nghĩ típ

Ta có: \(\frac{x-2019}{2018}+\frac{x-2018}{2017}=\frac{x-2017}{2016}+\frac{x-2016}{2015}\)

\(\Leftrightarrow\left(\frac{x-2019}{2018}+1\right)+\left(\frac{x-2018}{2017}+1\right)=\left(\frac{x-2017}{2016}+1\right)+\left(\frac{x-2016}{2015}+1\right)\)

\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}=\frac{x-1}{2016}+\frac{x-1}{2015}\)

\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}-\frac{x-1}{2016}-\frac{x-1}{2015}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)

\(\Leftrightarrow x-1=0\)( vì \(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\ne0\))

\(\Leftrightarrow x=1\)

Vạy x=1

\(\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2017}=\frac{x-1}{2021}+\frac{x-2}{2022}+\frac{x-3}{2023}\)

\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+\left(\frac{x+3}{2017}+1\right)=\left(\frac{x-1}{2021}+1\right)+\left(\frac{x-2}{2022}+1\right)+\left(\frac{x-3}{2023}+1\right)\)

\(\Leftrightarrow\left(\frac{x+1+2019}{2019}\right)+\left(\frac{x+2+2018}{2018}\right)+\left(\frac{x+3+2017}{2017}\right)=\left(\frac{x-1+2021}{2021}\right)+\left(\frac{x-2+2022}{2022}\right)+\left(\frac{x-3+2023}{2023}\right)\)

\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2021}+\frac{x+2020}{2022}+\frac{x+2020}{2023}\)

\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}-\frac{x+2020}{2021}-\frac{x+2020}{2022}-\frac{x+2020}{2023}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\right)=0\)

Vì \(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\ne0\)

=> x + 2020 = 0

=> x = -2020

            Bài làm :

Ta có :

\(\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2017}=\frac{x-1}{2021}+\frac{x-2}{2022}+\frac{x-3}{2023}\)

\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+\left(\frac{x+3}{2017}+1\right)=\left(\frac{x-1}{2021}+1\right)+\left(\frac{x-2}{2022}+1\right)+\left(\frac{x-3}{2023}+1\right)\)

\(\Leftrightarrow\left(\frac{x+1+2019}{2019}\right)+\left(\frac{x+2+2018}{2018}\right)+\left(\frac{x+3+2017}{2017}\right)=\left(\frac{x-1+2021}{2021}\right)+\left(\frac{x-2+2022}{2022}\right)+\left(\frac{x-3+2023}{2023}\right)\)

\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2021}+\frac{x+2020}{2022}+\frac{x+2020}{2023}\)

\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}-\frac{x+2020}{2021}-\frac{x+2020}{2022}-\frac{x+2020}{2023}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\right)=0\)

 \(\text{Vì : }\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\ne0\)

\(\Rightarrow x+2020=0\Leftrightarrow x=-2020\)

Vậy x=-2020

\(\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2017}=\frac{x-1}{2021}+\frac{x-2}{2022}+\frac{x-3}{2023}\)

\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+\left(\frac{x+3}{2017}+1\right)=\left(\frac{x-1}{2021}+1\right)+\left(\frac{x-2}{2022}+1\right)+\left(\frac{x-3}{2023}+1\right)\)

\(\Leftrightarrow\frac{x+1+2019}{2019}+\frac{x+2+2018}{2018}+\frac{x+3+2017}{2017}=\frac{x-1+2021}{2021}+\frac{x-2+2022}{2022}+\frac{x-3+2023}{2023}\)\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2021}+\frac{x+2020}{2022}+\frac{x+2020}{2023}\)

\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}-\frac{x+2020}{2021}-\frac{x+2020}{2022}-\frac{x+2020}{2023}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\right)=0\)

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)