We have: | x-1 |+|x-3|+|x-5|+|x-7|=|x-1|+|x-3|+|5-x|+|7-x|

\(\ge\)|x-1+x-3+5-x+7-x|=|8|=8

Mark ''='' occurs when \(\left(x-1\right)\left(x-3\right)\left(5-x\right)\left(7-x\right)\ge0\)\(\Rightarrow\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)\ge0\)

Case 1 :x-1 , x-3 , x-5 , x-7 \(\ge0\) so x=7 because if x>7 so (x-1)(x-3)(x-5)(x-7)>0(not satisfy)

Case 2 :x-1 , x-3\(\ge\)0 ; x-5 , x-7\(\le0\)\(\Rightarrow x-3\ge0;x-5\le0\)\(\Rightarrow5\ge x\ge3\)

Case 3 :x-1 , x-5 \(\ge\)0 ; x-3 , x-7\(\le\)0\(\Rightarrow x-5\ge0;x-3\le0\)\(\Rightarrow\left\{{}\begin{matrix}x\ge5\\x\le3\end{matrix}\right.\)(not satisfy)

Case 4 :x-1 , x-7 \(\ge0\) ; x-3 , x-5\(\le\)0\(\Rightarrow\left\{{}\begin{matrix}x\ge7\\x\le3\end{matrix}\right.\)(not satisfy)

Case 5 :x-1 , x-3 , x-5 , x-7\(\le\)0\(\Rightarrow x\le1\)

Case 6 :x-1 , x-3\(\le0\) ; x-5 , x-7\(\ge\)0\(\Rightarrow\left\{{}\begin{matrix}x\le1\\x\ge7\end{matrix}\right.\)(not satisfy)

Case 7 :x-1 , x-5\(\le\)0 ; x-3 , x-7\(\ge\)0\(\Rightarrow\left\{{}\begin{matrix}x\le1\\x\ge7\end{matrix}\right.\)(not satisfy)

Case 8 :x-1 , x-7\(\le0\) ; x-3 , x-5\(\ge0\)\(\Rightarrow\left\{{}\begin{matrix}x\le1\\x\ge5\end{matrix}\right.\)(not satisfy)

Retry x=7 we have |7-1|+|7-3|+|7-5|+|7-7|=12(not satisfy)

So max_x=5

We have: | x-1 |+|x-3|+|x-5|+|x-7|=|x-1|+|x-3|+|5-x|+|7-x|

≥

|x-1+x-3+5-x+7-x|=|8|=8

Mark ''='' occurs when (x−1)(x−3)(5−x)(7−x)≥0

⇒(x−1)(x−3)(x−5)(x−7)≥0

Case 1 :x-1 , x-3 , x-5 , x-7 ≥0

 so x=7 because if x>7 so (x-1)(x-3)(x-5)(x-7)>0(not satisfy)

Case 2 :x-1 , x-3≥

0 ; x-5 , x-7≤0⇒x−3≥0;x−5≤0⇒5≥x≥3

Case 3 :x-1 , x-5 ≥

0 ; x-3 , x-7≤0⇒x−5≥0;x−3≤0⇒{x≥5x≤3

(not satisfy)

Case 4 :x-1 , x-7 ≥0

 ; x-3 , x-5≤0⇒{x≥7x≤3

(not satisfy)

Case 5 :x-1 , x-3 , x-5 , x-7≤

0⇒x≤1

Case 6 :x-1 , x-3≤0

 ; x-5 , x-7≥0⇒{x≤1x≥7

(not satisfy)

Case 7 :x-1 , x-5≤

0 ; x-3 , x-7≥0⇒{x≤1x≥7

(not satisfy)

Case 8 :x-1 , x-7≤0

 ; x-3 , x-5≥0⇒{x≤1x≥5

(not satisfy)

Retry x=7 we have |7-1|+|7-3|+|7-5|+|7-7|=12(not satisfy)

So maxx=5

Các bạn lập bảng xét dấu: như vậy chỉ cần xét 5 trường hợp thôi:

1)  \(x\le1\) thì các biểu thức trong trị tuyệt đối đều âm ta có  VT=(1-x)+(3-x)+(5-x)+(7-x)=8

ta giải tìm x

Tương tự cho 4 trường hợp còn lại.