a)Dễ thấy \(|5x+3|\ge 0\)
\(\Rightarrow A=\left|5x+3\right|-2\ge-2\)
Khi \(x=-\dfrac{3}{5}\)
b)Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(B=\left|x+6\right|+\left|11+x\right|+6\)
\(=\left|x+6\right|+\left|-11-x\right|+6\)
\(\ge\left|x+6-11-x\right|+6=11\)
Khi \(-11\le x\le -6\)
\(A=\left|5x+3\right|-2\)
\(\left|5x+3\right|\ge0\forall x\in R\)
\(A=\left|5x+3\right|-2\ge-2\)
Dấu "=" xảy ra khi:
\(\left|5x+3\right|=0\Leftrightarrow x=-\dfrac{3}{5}\)
\(B=\left|x+6\right|+\left|11+x\right|\)
\(B=\left|x+6\right|+\left|-11-x\right|\)
Áp dụng bđt:
\(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(B\ge\left|x+6-11-x\right|\)
\(B\ge5\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+6\ge0\Rightarrow x\ge-6\\-11-x\ge0\Rightarrow x\le-11\end{matrix}\right.\\\left\{{}\begin{matrix}x+6< 0\Rightarrow x< -6\\-11-x< 0\Rightarrow x>-11\end{matrix}\right.\end{matrix}\right.\)
Vậy \(-11< x< -6\)
