NM
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
RR
16 tháng 5 2018
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
S
29 tháng 12 2020
\(\dfrac{x-1}{2019}+\dfrac{x-2}{2018}+\dfrac{x-3}{2017}=3\)
\(\Leftrightarrow\left(\dfrac{x-1}{2019}-1\right)+\left(\dfrac{x-2}{2018}-1\right)+\left(\dfrac{x-3}{2017}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-1-2019}{2019}+\dfrac{x-2-2018}{2018}+\dfrac{x-3-2017}{2017}=0\)
\(\Leftrightarrow\dfrac{x-2020}{2019}+\dfrac{x-2020}{2018}+\dfrac{x-2020}{2017}=0\)
\(\Leftrightarrow\left(x-2020\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}+\dfrac{1}{2017}\right)=0\)
Vi \(\dfrac{1}{2019}+\dfrac{1}{2018}+\dfrac{1}{2017}\ne0\)
nên \(x-2020=0\)
\(\Leftrightarrow x=2020\)
Vậy ...
GA
2
29 tháng 1 2020
\(\left|2017-x\right|+\left|2018-x\right|+\left|2019-x\right|=2\left(1\right)\)
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 \(x=2018\)
4 tháng 5 2018
=>|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}
NN
3
17 tháng 9 2020
\(\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2017}=\frac{x-1}{2021}+\frac{x-2}{2022}+\frac{x-3}{2023}\)
\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+\left(\frac{x+3}{2017}+1\right)=\left(\frac{x-1}{2021}+1\right)+\left(\frac{x-2}{2022}+1\right)+\left(\frac{x-3}{2023}+1\right)\)
\(\Leftrightarrow\left(\frac{x+1+2019}{2019}\right)+\left(\frac{x+2+2018}{2018}\right)+\left(\frac{x+3+2017}{2017}\right)=\left(\frac{x-1+2021}{2021}\right)+\left(\frac{x-2+2022}{2022}\right)+\left(\frac{x-3+2023}{2023}\right)\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2021}+\frac{x+2020}{2022}+\frac{x+2020}{2023}\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}-\frac{x+2020}{2021}-\frac{x+2020}{2022}-\frac{x+2020}{2023}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\right)=0\)
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
B
17 tháng 9 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
TL
1
15 tháng 9 2019
\(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}+\frac{x+4}{2017}=-4\)
=> \(\left[\frac{x+1}{2020}+1\right]+\left[\frac{x+2}{2019}+1\right]+\left[\frac{x+3}{2018}+1\right]+\left[\frac{x+4}{2017}+1\right]=-4\)
=> \(\left[\frac{x+1}{2020}+\frac{2020}{2020}\right]+\left[\frac{x+2}{2019}+\frac{2019}{2019}\right]+\left[\frac{x+3}{2018}+\frac{2018}{2018}\right]+\left[\frac{x+4}{2017}+\frac{2017}{2017}\right]=-4\)
=> \(\frac{x+2021}{2020}+\frac{x+2021}{2019}+\frac{x+2021}{2018}+\frac{x+2021}{2017}=-4\)
=> \(\left[x+2021\right]\left[\frac{1}{2000}+\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}\right]=-4\)
Do \(\frac{1}{2020}>\frac{1}{2019}>\frac{1}{2018}>\frac{1}{2017}\)nên \(\frac{1}{2000}+\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}\ne0\)
Do đó : x + 2021 = -4 => x = -4 - 2021 = -2025
3 tháng 5 2019
x−1/2019+x−2/2018=x−3/2017+x−4/2016(đề có thiếu không bạn??)
⇔(x−1/2019−1)+(x−2/2018−1)=(x−3/2017−1)+(x−4/2016−1)
⇔x−2020/2019+x−2020/2018=x−2020/2017+x−2020/2016
⇔x−2020/2019+x−2020/2018−x−2020/2017−x−2020/2016
⇔(x−2020)(1/2019+1/2018−1/2017−1/2016)=0
Mà 1/2019+1/2018−1/2017−1/2016≠0
⇔x−2020=0
⇔x=2020
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