a) \(A=\left|x+2\right|+\left|x-3\right|\)
\(\Leftrightarrow A=\left|x+2\right|+\left|3-x\right|\)
Ta có: \(\left|x+2\right|+\left|3-x\right|\ge\left|x+2+3-x\right|\)
\(\Rightarrow\left|x+2\right|+\left|3-x\right|\ge5\)
\(\Leftrightarrow\left(x+2\right)\left(3-x\right)\ge0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)\le0\)
Mà \(x+2>x-3\)
\(\Rightarrow\left\{{}\begin{matrix}x+2\ge0\\x-3\le0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ge-2\\x\le3\end{matrix}\right.\)
Vậy \(Min_A=5\) đạt được \(\Leftrightarrow-2\le x\le3\)
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