Question

tìm GTNN của biểu thức: A= /x-1/+/x-2/+/x-3/

Answer 1

                                                          Bài giải

\(A=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\)

\(A=\left|x-1\right|+\left|2-x\right|+\left|x-3\right|\ge\left|x-1+2-x\right|+\left|x-3\right|=\left|1\right|+\left|x-3\right|=1+\left|x-3\right|\ge1\)

Dấu " = " xảy ra khi \(1\le x\le2\)

Vậy Min A = 1 khi \(1\le x\le2\)

Answer 2

Nhầm Min là 2 khi x = 2 nha !

Answer 3

Ta có: \(\left|x-2\right|\ge0\),

           \(\left|x-1\right|+\left|x-3\right|=\left|x-1\right|+\left|3-x\right|\ge\left|x-1+3-x\right|=2\)(theo BĐT |a|+|b| lớn hơn or bằng |a+b|)

=> GTNN A=0+2=2

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-2=0\\\left(x-1\right)\left(3-x\right)\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\1\le x\le3\end{cases}\Leftrightarrow}x=2}\)

       

Answer 4

                                                              Bài giải

Ta có : \(A=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\)

\(A=\left|x-1\right|+\left|2-x\right|+\left|x-3\right|\ge\left|x-1+2-x\right|+\left|x-3\right|=\left|-1\right|+\left|x-3\right|=1+\left|x-3\right|\)

Dấu " = " xảy ra khi \(1\le x\le2\)\(\Rightarrow\) \(x=2\) \(\Rightarrow\) \(A\ge1+1=2\)

Vậy Min A = 2 khi x = 2

Answer 5

\(A=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=\left|x-1\right|+\left|x-3\right|+\left|x-2\right|\)

\(=\left|x-1\right|+\left|3-x\right|+\left|x-2\right|\ge\left|x-1+3-x\right|+\left|x-2\right|\)

\(=\left|2\right|+\left|x-2\right|=2+\left|x-2\right|\ge2\)( vì \(\left|x-2\right|\ge0\))

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)\left(3-x\right)\ge0\\x-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(x-1\right)\left(3-x\right)\ge0\\x=2\end{cases}}\)(1)

Xét \(\left(x-1\right)\left(3-x\right)\ge0\)

TH1: \(\hept{\begin{cases}x-1< 0\\3-x< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 1\\3< x\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 1\\x>3\end{cases}}\)( vô lý )

TH2: \(\hept{\begin{cases}x-1\ge0\\3-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\3\ge x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\3\le x\end{cases}}\Leftrightarrow1\le x\le3\)(2)

Từ (1) và (2) ta có: \(\hept{\begin{cases}1\le x\le3\\x=2\end{cases}}\Leftrightarrow x=2\)

Vậy \(minA=2\Leftrightarrow x=2\)