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ND
1
DD
Đoàn Đức Hà
Giáo viên
14 tháng 7 2021
\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+...+\left|x-2020\right|\)
\(=\left(\left|x-1\right|+\left|2020-x\right|\right)+\left(\left|x-2\right|+\left|2019-x\right|\right)+...+\left(\left|x-1010\right|+\left|1011-x\right|\right)\)
\(\ge\left|x-1+2020-x\right|+\left|x-2+2019-x\right|+...+\left|x-1010+1011-x\right|\)
\(=2019+2017+...+1\)
\(=\frac{\left(2019+1\right).\left[\left(2019-1\right)\div2+1\right]}{2}=1020100\)
Dấu \(=\)khi \(\hept{\begin{cases}\left(x-1\right)\left(2020-x\right)\ge0\\...\\\left(x-1010\right)\left(1011-x\right)\ge0\end{cases}}\Leftrightarrow1010\le x\le1011\).
14 tháng 7 2021
`|x-1|+|x-2|+|x-3|+....+|x-2020|`
`=(|x-1|+|x-2020|)+(|x-2|+|x-2019|)+....+(|x-1000|+|x-1001|)`
Áp dụng bđt `|A|+|B|>=|A+B|`
`=>|x-1|+|x-2020|=|x-1|+|2020-x|>=|x-1+2020-x|=2019`
Tương tự:
`|x-2|+|x-2019|>=2017`
`.................................`
`|x-1000|+|x-1001|>=1`
`=>|x-1|+|x-2|+|x-3|+....+|x-2020|>=2019+2017+....+1`
`=>|x-1|+|x-2|+|x-3|+....+|x-2020|>=((2019+1).2019)/2=2039190`
Dấu "=" xảy ra khi `{((x-1)(2020-x)>=0),((x-2)(2019-x)>=0),(.........),((x-1000)(1001-x)>=0):}`
`<=>{((x-1)(x-2020)<=0),((x-2)(x-2019)<=0),(.........),((x-1000)(x-1001)<=0):}`
`<=>{(1<=x<=2020),(2<=x<=2019),(.........),(1000<=x<=1001):}`
`<=>1000<=x<=1001`
NN
1
DL
1 tháng 3 2022
|x−1|+|x−2|+|x−3|+...+|x−2020||x−1|+|x−2|+|x−3|+...+|x−2020|
=(|x−1|+|2020−x|)+(|x−2|+|2019−x|)+...+(|x−1010|+|1011−x|)=(|x−1|+|2020−x|)+(|x−2|+|2019−x|)+...+(|x−1010|+|1011−x|)
≥|x−1+2020−x|+|x−2+2019−x|+...+|x−1010+1011−x|≥|x−1+2020−x|+|x−2+2019−x|+...+|x−1010+1011−x|
=2019+2017+...+1=2019+2017+...+1
=(2019+1).[(2019−1)÷2+1]2=1020100=(2019+1).[(2019−1)÷2+1]2=1020100
Dấu ==khi \hept⎧⎨⎩(x−1)(2020−x)≥0...(x−1010)(1011−x)≥0⇔1010≤x≤1011\hept{(x−1)(2020−x)≥0...(x−1010)(1011−x)≥0⇔1010≤x≤1011.
NM
1
14 tháng 12 2021
\(N=\left(\left|x-1\right|+\left|2020-x\right|\right)+\left(\left|x-2\right|+\left|2019-x\right|\right)+...+\left(\left|x-1010\right|+\left|1011-x\right|\right)\\ N\ge\left|x-1+2020-x\right|+\left|x-2+2019-x\right|+...+\left|x-1010+1011-x\right|\\ N\ge2019+2017+...+1=\dfrac{\left(2019+1\right)\left[\left(2019-1\right):2+1\right]}{2}=1020100\\ N_{min}=1020100\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(2020-x\right)\ge0\\...\\\left(x-1010\right)\left(1011-x\right)\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1\le x\le2020\\...\\1010\le x\le1011\end{matrix}\right.\Leftrightarrow1010\le x\le1011\)
8 tháng 3 2020
1, Ta có: \(\left(x-y\right)^6+|47-x|+3^3\ge0+0+9=9\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}x-y=0\\47-x=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=47\\y=47\end{cases}}\)
2, Ta có: \(\left(x+5\right)^2+\left(y-9\right)^2+2020\ge0+0+2020=2020\)
Dấu "'=" xảy ra khi \(\hept{\begin{cases}x+5=0\\y-9=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\y=9\end{cases}}}\)
TL
0
KM
0
DK
4
14 tháng 2 2020
Ta có: \(A=\left|x-2018\right|+\left|2019-x\right|+\left|x-2020\right|\)
\(A=\left(\left|x-2018\right|+\left|2020-x\right|\right)+\left|2019-x\right|\)
\(\Rightarrow A\ge\left|x-2018+2020-x\right|+\left|2019-x\right|=2+\left|2019-x\right|\)
Dấu "=" xảy ra <=> \(\left(x-2018\right)\left(2020-x\right)\ge0\)
\(\Rightarrow\left(x-2018\right)\left(x-2020\right)\le0\)
\(\Rightarrow\hept{\begin{cases}x-2018\ge0\\x-2020\le0\end{cases}\Rightarrow\hept{\begin{cases}x\ge2018\\x\le2020\end{cases}\Rightarrow}2018\le x\le2020}\)
Và \(\left|2019-x\right|\ge0\), Min (A) = 2 <=> |2019-x| = 0 <=> x= 2019