Lời giải:
Ta có:
\(H=|x-3|+|4+x|=|3-x|+|4+x|\geq |3-x+4+x|\)
\(\Leftrightarrow H\geq |7|=7\)
Vậy \(H_{\min}=7\)
Dấu bằng xảy ra khi \((3-x)(x+4)\geq 0\Leftrightarrow -4\leq x\leq 3\)
\(H=\left|x-3\right|+\left|4+x\right|=\left|3-x\right|+\left|4+x\right|\ge\left|3-x+4+x\right|=7\)
Dấu "=" xảy ra khi: \(-4\le x\le3\)
\(H=\left|x-3\right|+\left|4+x\right|\)
\(\Leftrightarrow H=\left|3-x\right|+\left|4+x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(H=\left|3-x\right|+\left|4+x\right|\ge\left|3-x+4+x\right|=7\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}3-x\ge0\\4+x\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge3\\x\le4\end{matrix}\right.\) \(\Rightarrow3\le x\le4\)
Vậy GTNN của \(H=7\) khi \(3\le x\le4\)
Ta có:
\(\left\{{}\begin{matrix}\left|x-3\right|=\left|3-x\right|\ge3-x\\\left|4+x\right|\ge4+x\end{matrix}\right.\)
\(\Rightarrow\left|x-3\right|+\left|4+x\right|\ge\left(3-x\right)+\left(4+x\right)\)
\(\Rightarrow H\ge7\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}3-x\ge0\\4+x\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le3-0\\x\ge0-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le3\\x\ge-4\end{matrix}\right.\)
Vậy MinH = 7 \(\Leftrightarrow\left\{{}\begin{matrix}x\le3\\x\ge-4\end{matrix}\right.\)
