\(P=\left|x+3\right|+\left|\frac{1}{6}-x\right|+\left|x-\frac{1}{6}\right|+\left|1-x\right|+\frac{5}{3}\left|6x-1\right|\)
\(P\ge\left|x+3+\frac{1}{6}-x\right|+\left|x-\frac{1}{6}+1-x\right|+\frac{5}{3}\left|6x-1\right|\)
\(P\ge4+\frac{5}{3}\left|6x-1\right|\ge4\)
Dấu "=" xảy ra khi \(x=\frac{1}{6}\Rightarrow P=1.6=6\)