Tìm x |2017-|+|2018-x|+|2019-x|=2
|2017-x|+|2018-x|+|2019-x|=2.Tìm x
Vì |2017-x|>=x-2017
|2018-x|>=0
|2019-x|>=2019-x
=>|2017-x|+|2018-x|+|2019-x|>=2
Dấu = xảy ra <=> x>2017
x=2018
x<2019
Vậy x=2018
Tìm x biết: |2017-x| + |2018-x| + |2019-x| =2
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
tìm x biết 2017 x |+| 2018 x |+| 2019 x |= 2
|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
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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
tìm x biết
x-1/2019+x-2/2018+x-3/2017=3
tìm x biết
x-1/2019+x-2/2018+x-3/2017=3
\(\dfrac{x-1}{2019}+\dfrac{x-2}{2018}+\dfrac{x-3}{2017}=3\)
\(\Leftrightarrow\left(\dfrac{x-1}{2019}-1\right)+\left(\dfrac{x-2}{2018}-1\right)+\left(\dfrac{x-3}{2017}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-1-2019}{2019}+\dfrac{x-2-2018}{2018}+\dfrac{x-3-2017}{2017}=0\)
\(\Leftrightarrow\dfrac{x-2020}{2019}+\dfrac{x-2020}{2018}+\dfrac{x-2020}{2017}=0\)
\(\Leftrightarrow\left(x-2020\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}+\dfrac{1}{2017}\right)=0\)
Vi \(\dfrac{1}{2019}+\dfrac{1}{2018}+\dfrac{1}{2017}\ne0\)
nên \(x-2020=0\)
\(\Leftrightarrow x=2020\)
Vậy ...
( 2019 . x - 2019 ) . (2019 . 2018 -2017 )
tìm x
Tìm x,biết :|2017-x|+|2018-x|+|2019-x|=2
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 \(x=2018\)
Tìm x biết :
[1/2 + 1/3 + .......+ 1/2019]x = 2018/1 + 2017/2 + .......+ 1/2018
1. Tìm các số nguyên x , y thỏa mãn :
2017 + x2018 + 20172018 = y2018 + y2019 + 20182019