\(A=\left|x+1\right|+\left|x+2\right|+\left|x+3\right|\)
\(A=\left|x+1\right|+\left|x+3\right|+\left|x+2\right|\)
\(A=\left|x+1\right|+\left|-x-3\right|+\left|x+2\right|\)
\(A\ge\left|x+1-x-3\right|+\left|x+2\right|\)
\(A\ge2+\left|x+2\right|\)
Vì \(\left|x+2\right|\ge0\) nên \(A\ge2\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}x+1\le0\\x+2=0\\x+3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le-1\\x=-2\\x\ge-3\end{matrix}\right.\Leftrightarrow x=-2\)
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