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 ...

bạn viết thiếu đè bạn êy

Đề là vậy phải không : | 2017 - x | + | 2018 - x | + | 2019 - x | = 2

Nếu x ≤ 2017 thì | 2017 - x | = 2017 - x 

                              | 2018 - x | = 2018 - x

                              | 2019 - x | = 2019 - x

Pt (=) 2017-x+2018-x+2019-x = 2

(=) -3x + 6054 = 2

(=) 3x = 6052

(=) x = 6052/3 ( loại, vì > 2017 )

Nếu 2017 < x < 2018 thì | 2017 - x | = x - 2017 ; | 2018 - x | = 2018 - x ; | 2019 - x | = 2019 - x

Pt (=) x-2017+2018-x+2019-x = 2

(=) x = -2018 ( loại )

Nếu 2018 ≤ x ≤ 2019 thì | 2017-x| = x-2017 ; | 2018-x| = x-2018 ; | 2019-x | = 2019-x

Pt (=) x-2017+x-2018+2019-x = 2

(=) x = 2018 ( TM )

Nếu x > 2019 thì | 2017-x | = x-2017 ; | 2018-x | = x-2018  ; | 2019-x | = x-2019

Pt(=) x-2017+x-2018+x-2019 = 2

(=) 3x = 6056

(=) x = 6056/3 ( loại )

x = 2018 hoặc x = +2018            chất vl

\(\left|2017-x\right|+\left|2018-x\right|+\left|2019-x\right|=2\left(1\right)\)

TH1: \(x\le2017\)

\(\left(1\right)\Leftrightarrow2017-x+2018-x+2019-x=2\)

\(\Rightarrow6054-3x=2\)

\(\Rightarrow3x=6052\)

\(\Rightarrow x=\frac{6052}{3}\)(loại)

TH2: \(2017< x\le2018\)

\(\left(1\right)\Leftrightarrow x-2017+2018-x+2019-x=2\)

\(\Rightarrow2020-x=2\)

\(\Rightarrow x=2018\)(thỏa mãn điều kiện)

TH3: \(2018< x\le2019\)

\(\left(1\right)\Leftrightarrow x-2017+x-2018+2019-x=2\)

\(\Rightarrow x-2016=2\)

\(\Rightarrow x=2018\)(thỏa mãn điều kiện)

TH4: \(x>2019\)

\(\left(1\right)\Leftrightarrow x-2017+x-2018+x-2019=2\)

\(\Rightarrow3x=6056\)

\(\Rightarrow x=\frac{6056}{3}\)(loại)

Vậy \(x=2018\)

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