a) Ta có: \(\left|15x-1\right|\ge37\)
\(\Leftrightarrow\left[{}\begin{matrix}15x-1\ge37\\15x-1\le-37\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}15x\ge38\\15x\le-36\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge\frac{38}{15}\\x\le\frac{-12}{5}\end{matrix}\right.\)
Vậy: S={\(x\in R\)|\(\left[{}\begin{matrix}x\ge\frac{38}{15}\\x\le\frac{-12}{5}\end{matrix}\right.\)}
b) Ta có: |x+1|+|x+3|=3x(1)
Trường hợp 1: x<-3
(1)\(\Leftrightarrow-x-1-x-3=3x\)
\(\Leftrightarrow-2x-4-3x=0\)
\(\Leftrightarrow-5x-4=0\)
\(\Leftrightarrow-5x=4\)
hay \(x=\frac{-4}{5}\)(loại)
Trường hợp 2: \(-3\le x< -1\)
(1)\(\Leftrightarrow x+3-x-1=3x\)
\(\Leftrightarrow3x=2\)
hay \(x=\frac{2}{3}\)(loại)
Trường hợp 3: \(x\ge-1\)
(1)\(\Leftrightarrow x+1+x+3=3x\)
\(\Leftrightarrow2x+4-3x=0\)
\(\Leftrightarrow-x+4=0\)
hay x=4(nhận)
Vậy: S={4}
d) Ta có: |x-1|+|x-4|=4x(2)
Trường hợp 1: x<1
(2)\(\Leftrightarrow1-x+4-x=4x\)
\(\Leftrightarrow5-2x-4x=0\)
\(\Leftrightarrow-6x+5=0\)
\(\Leftrightarrow-6x=-5\)
hay \(x=\frac{5}{6}\)(nhận)
Trường hợp 2: \(1\le x< 4\)
(2)\(\Leftrightarrow x-1+4-x=4x\)
\(\Leftrightarrow4x=3\)
hay \(x=\frac{3}{4}\)(nhận)
Trường hợp 2: \(x\ge4\)
(2)\(\Leftrightarrow x-1+x-4=4x\)
\(\Leftrightarrow2x-5-4x=0\)
\(\Leftrightarrow-2x-5=0\)
\(\Leftrightarrow-2x=5\)
hay \(x=\frac{-5}{2}\)(loại)
Vậy: \(S=\left\{\frac{5}{6}\right\}\)
