\(A=\left|x-2011\right|+\left|x-2012\right|+\left|x-2013\right|+\left|x-2014\right|+\left|x-2015\right|\)

\(=\left(\left|x-2011\right|+\left|2015-x\right|\right)+\left(\left|x-2012\right|+\left|2014-x\right|\right)+\left|x-2013\right|\)

Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) , dấu "=" xảy ra khi a,b cùng dấu. Ta có : \(\left|x-2011\right|+\left|2015-x\right|\ge\left|x-2011+2015-x\right|=4\)

\(\left|x-2012\right|+\left|2014-x\right|\ge\left|x-2012+2014-x\right|=2\)

\(\left|x-2013\right|\ge0\)

\(\Rightarrow A\ge4+2+0=6\)

Dấu "=" xảy ra khi \(\begin{cases}2011\le x\le2015\\2012\le x\le2014\\x=2013\end{cases}\) \(\Leftrightarrow x=2013\)

Vậy A đạt giá trị nhỏ nhất bằng 6 tại x = 2013

x=2013

\(A=\left|x-2011\right|+\left|x-2012\right|+\left|x-2013\right|+\left|x-2014\right|+\left|x-2015\right|\)

\(=\left(\left|x-2011\right|+\left|x-2015\right|\right)+\left(\left|x-2012\right|+\left|x-2014\right|\right)+\left|x-2013\right|\)

Đặt \(B=\left|x-2011\right|+\left|x-2015\right|\)

\(=\left|x-2011\right|+\left|2015-x\right|\ge\left|x-2011+2015-x\right|=4\left(1\right)\)

Dấu"=" xảy ra \(\Leftrightarrow\left(x-2011\right)\left(2015-x\right)\ge0\)

\(\Leftrightarrow\hept{\begin{cases}x-2011\ge0\\2015-x\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x-2011< 0\\2015-x< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge2011\\x\le2015\end{cases}}\)hoặc \(\hept{\begin{cases}x< 2011\\x>2015\end{cases}\left(loai\right)}\)

\(\Leftrightarrow2011\le x\le2015\)

Đặt \(C=\left|x-2012\right|+\left|x-2014\right|\)

\(=\left|x-2012\right|+\left|2014-x\right|\ge\left|x-2012+2014-x\right|=2\left(2\right)\)

Dấu"="xảy ra \(\Leftrightarrow\left(x-2012\right)\left(2014-x\right)\ge0\)

\(\Leftrightarrow\hept{\begin{cases}x-2012\ge0\\2014-x\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x-2012< 0\\2014-x< 0\end{cases}}\) 

\(\Leftrightarrow\hept{\begin{cases}x\ge2012\\x\le2014\end{cases}}\)hoặc\(\hept{\begin{cases}x< 2012\\x>2014\end{cases}\left(loai\right)}\)

\(\Leftrightarrow2012\le x\le2014\)

Ta có: \(\left|x-2013\right|\ge0;\forall x\left(3\right)\)

Dấu"="Xảy ra \(\Leftrightarrow\left|x-2013\right|=0\)

                      \(\Leftrightarrow x=2013\)

Từ (1),(2) và (3) \(\Rightarrow B+C+\left|x-2013\right|\ge6\)

Hay \(A\ge6\)

Dấu"="xảy ra \(\Leftrightarrow\hept{\begin{cases}2011\le x\le2015\\2012\le x\le2014\\x=2013\end{cases}}\)\(\Leftrightarrow x=2013\)

Vậy \(A_{min}=6\Leftrightarrow x=2013\)

=4050151.001

2012×2013+2011/2014×2013-2015=2012×2013+2011/(2012+2)×2013-2015=2012×2013+2011/2012×2013+2×2015=2012×2013+2011/2012×2013+4026-2015=2012×2013+2011/2012×2013+2011=1

\(\frac{x+4}{2010}+\frac{x+3}{2011}=\frac{x+2}{2012}+\frac{x+1}{2013}\)

\(\Leftrightarrow\left(\frac{x+4}{2010}+1\right)+\left(\frac{x+3}{2011}+1\right)=\left(\frac{x+2}{2012}+1\right)+\left(\frac{x+1}{2013}+1\right)\)

\(\Leftrightarrow\frac{x+2014}{2010}+\frac{x+2014}{2011}=\frac{x+2014}{2012}+\frac{x+2014}{2013}\)

\(\Leftrightarrow\frac{x+2014}{2010}+\frac{x+2014}{2011}-\frac{x+2014}{2012}-\frac{x+2014}{2013}=0\)

\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)=0\)

\(\Leftrightarrow x+2014=0\)

\(\Leftrightarrow x=-2014\)

V...

cong 1 vao tung hang tu (vd 3/2013) roi dc tong 2016

\(\frac{x+3}{2013}+1+\)\(\frac{x+4}{2012}+1+\frac{x+5}{2011}+1\)=\(\frac{x+1}{2015}+1+\frac{x+2}{2014}+1+\frac{x}{2016}+1\)

 \(\Rightarrow\frac{x+2016}{2013}+\frac{x+2016}{2012}+\frac{x+2016}{2011}=\frac{x+2016}{2014}+\frac{x+2016}{2016}\)

\(\Rightarrow\left(2016+x\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2016}=0\right)\)

Vì 1/2016+...+1/2011>0 nên (x+2016)=0

suy ra x= -2016

nếu đúng xin kết bạn

ch

toán lớp 1 khó thế

rối loạn mắt rồi hu hu

Bài làm:

\(x=\frac{2012}{2013}+\frac{2013}{2014}+\frac{2014}{2015}\)

\(x=\frac{2012.2014.2015+2013.2013.2015+2014.2013.2014}{2013.2014.2015}\)

\(x=\frac{8165118520+8165120535+8165122548}{8169176730}\)

\(x=\frac{24495361603}{8169176730}\)

Học tốt!

>: đúng vậy ... mỗi cách đấy ! , nhưng @Đăng ... chép sai đề kìa 

\(x=\frac{2012}{2013}+\frac{2013}{2014}+\frac{2015}{2016}\)

\(=\frac{2012.2014.2016}{2013.2014.2016}+\frac{2013.2012.2016}{2014.2013.2016}+\frac{2015.2013.2014}{2016.2014.2013}\)

\(=\frac{2012.2014.2016+2013.2012.2016+2015.2013.2014}{2013.2014.2016}\)

\(=\frac{8169170688+8165114496+8169176790}{8173290912}\)

Để thế này cx đc >: chả ảnh hưởng j cả, mà sợ đáp án sai vì số dài ... xem lại hộ nhé !