1.
a) \(\left|3x+6\right|+5x-1=0\)
\(\Rightarrow\left|3x+6\right|=1-5x\)
+) TH1: \(3x+6\ge0\Rightarrow3x\ge-6\Rightarrow x\ge-2\)
Ta có: \(3x+6=1-5x\)
\(\Rightarrow3x+5x=-6+1\)
\(\Rightarrow8x=-5\)
\(\Rightarrow x=\dfrac{-5}{8}\) (t/m)
+) TH2: \(3x+6< 0\Rightarrow3x< -6\Rightarrow x< -2\)
Khi đó \(-3x-6=1-5x\)
\(\Rightarrow-3x+5x=6+1\)
\(\Rightarrow2x=7\)
\(\Rightarrow x=\dfrac{7}{2}\) (loại)
Vậy \(x=\dfrac{-5}{8}\)
b) Vì \(\left|x+1\right|\ge x+1\forall x\)
\(\left|6-x\right|\ge6-x\forall x\)
\(\Rightarrow\left|x+1\right|+\left|6-x\right|\ge x+1+6-x\)
\(\Rightarrow\left|x+1\right|+\left|6-x\right|\ge7\)
Dấu \("="\) xảy ra khi
\(\left\{{}\begin{matrix}x+1\ge0\\6-x\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge-1\\x\le6\end{matrix}\right.\) \(\Rightarrow-1\le x\le6\)
Vậy \(-1\le x\le6.\)
