Question

[ღ Rain...]

Tìm giá trị nhỏ nhất của biểu thức:

#Hỏi cộng đồng OLM #Toán lớp 7

Answer 1

[Nguyễn Thanh Hằng]

Với mọi x ta có :

\(\left|x-2011\right|=\left|2011-x\right|\)

\(\Leftrightarrow\left|x-2018\right|+\left|x-2011\right|=\left|x-2018\right|+\left|2011-x\right|\)

\(\Leftrightarrow\left|x-2018\right|+\left|2011-x\right|\ge\left|\left(x-2018\right)+\left(2011-x\right)\right|\)

\(\Leftrightarrow\left|x-2018\right|+\left|2011-x\right|\ge7\)

Mà \(\left|y-2010\right|\ge0\)

\(\left|x-2009\right|\ge0\)

\(\Leftrightarrow\left|x-2008\right|+\left|x-2009\right|+\left|x-2011\right|+\left|y-2010\right|\ge7\)

\(\Leftrightarrow A\ge2015\)

Dấu "=" xảy ra khi :

\(\left\{{}\begin{matrix}\left(x-2008\right)\left(2011-x\right)\ge0\left(1\right)\\\left|y-2010\right|=0\left(2\right)\\\left|x-2009\right|=0\left(3\right)\end{matrix}\right.\)

Từ \(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2008\ge0\\2011-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2008\le0\\2011-x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2008\\2011\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2008\\2011\le x\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2008\le x\le2011\\x\in\varnothing\end{matrix}\right.\)

\(\Leftrightarrow2008\ge x\ge2011\left(I\right)\)

Từ \(\left(2\right)\Leftrightarrow y-2010=0\)

\(\Leftrightarrow y=2010\left(II\right)\)

Từ \(\left(3\right)\Leftrightarrow x-2009=0\)

\(\Leftrightarrow x=2009\left(III\right)\)

Từ \(\left(I\right)+\left(II\right)+\left(III\right)\Leftrightarrow A_{Min}=2015\Leftrightarrow x=2009;y=2010\)