\(A=\left|x+3\right|+ \left|x+2\right|+7\)
\(A=\left|x+3\right|+\left|-x-2\right|+7\)
Đặt:
\(C=\left|x+3\right|+\left|-x-2\right|\)
\(C\ge\left|x+3-x-2\right|\)
\(C\ge1\)
\(\Rightarrow A\ge8\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0\Rightarrow x\ge-3\\x+2\ge0\Rightarrow x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x+3< 0\Rightarrow x< -3\\x+2< 0\Rightarrow x< -2\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x< -2\) hoặc \(x\ge-3\)
\(B=\left|x-6\right|+\left|x+1\right|+5\)
\(B=\left|x-6\right|+\left|-x-1\right|+5\)
Đặt:
\(D=\left|x-6\right|+\left|-x-1\right|\)
\(D\ge\left|x-6-x-1\right|\)
\(D\ge7\)
\(\Rightarrow B\ge12\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-6\ge0\Rightarrow x\le6\\x+1\ge0\Rightarrow x\ge-1\end{matrix}\right.\\\left\{{}\begin{matrix}x-6< 0\Rightarrow x>6\\x+1< 0\Rightarrow x< -1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-1\le x\le6\)
Vậy...
