\(\left|x-5\right|+\left|x-1\right|=\left|5-x\right|+\left|x-1\right|\ge\left|5-x+x-1\right|=\left|4\right|=4\\ \left|y+5\right|\ge0\\ \Leftrightarrow\left|y+5\right|+3\ge3\\ \Leftrightarrow\dfrac{12}{\left|y+5\right|+3}\le\dfrac{12}{3}=4\\ VT\ge4;VP\le4\\ \Rightarrow\text{Dấu "=" phải xảy ra }\Leftrightarrow\left\{{}\begin{matrix}\left(5-x\right)\left(x-1\right)\ge0\\y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1\le x\le5\\y=-5\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}\left|x-5\right|+\left|x-1\right|=\left|5-x\right|+\left|x-1\right|\ge\left|5-x+x-1\right|\ge\left|4\right|=4\\\dfrac{12}{\left|y+5\right|+3}\le\dfrac{12}{3}=4\end{matrix}\right..\)
\(\Rightarrow\left|x-5\right|+x-1=\dfrac{12}{\left|y+5\right|+3}=4.\)
\(\Rightarrow\left\{{}\begin{matrix}1< x< 5\\y=-5\end{matrix}\right..\)
Vậy..........
|x−5|+|x−1|=|5−x|+|x−1|≥|5−x+x−1|=|4|=4|y+5|≥0⇔|y+5|+3≥3⇔12|y+5|+3≤123=4VT≥4;VP≤4⇒Dấu "=" phải xảy ra ⇔{(5−x)(x−1)≥0y+5=0⇔{1≤x≤5y=−5
