a) \(\left|4x+3\right|-x=15\)\\
\(\Rightarrow\left|4x+3\right|=15+x.\)
\(\Rightarrow\left[{}\begin{matrix}4x+3=15+x\\4x+3=-15-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x-x=15-3\\4x+x=-15-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=12\\5x=-18\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{18}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{4;-\dfrac{18}{5}\right\}.\)
b) \(\left|3x-2\right|-x>1\)
\(\Rightarrow\left|3x-2\right|>1+x.\)
\(\Rightarrow\left[{}\begin{matrix}3x-2>1+x\\3x-2< -1-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-x>1+2\\3x+x< -1+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x>3\\4x< 1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< \dfrac{1}{4}\end{matrix}\right.\Rightarrow\dfrac{1}{4}< x< \dfrac{3}{2}.\)
Vậy \(\dfrac{1}{4}< x< \dfrac{3}{2}\)
c) \(\left|2x+3\right|\le5\)
\(\Rightarrow\left[{}\begin{matrix}2x+3\le5\\2x+3\ge-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x\le2\\2x\ge-8\end{matrix}\right.
\(\Rightarrow\left[{}\begin{matrix}x\le1\\x\ge-4\end{matrix}\right.\Rightarrow-4\le x\le1.\)
Vậy \(-4\le x\le1\)
a) \(\left|4x+3\right|-x=15\)
\(\Rightarrow\left|4x+3\right|=15+x.\)
\(\Rightarrow\left[{}\begin{matrix}4x+3=15+x\\4x+3=-15-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x-x=15-3\\4x+x=-15-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=12\\5x=-18\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{18}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{4;-\dfrac{18}{5}\right\}.\)
b) \(\left|3x-2\right|-x>1\)
\(\Rightarrow\left|3x-2\right|>1+x.\)
\(\Rightarrow\left[{}\begin{matrix}3x-2>1+x\\3x-2< -1-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-x>1+2\\3x+x< -1+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x>3\\4x< 1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< \dfrac{1}{4}\end{matrix}\right.\Rightarrow\dfrac{1}{4}< x< \dfrac{3}{2}.\)
Vậy \(\dfrac{1}{4}< x< \dfrac{3}{2}\)
c) \(\left|2x+3\right|\le5\)
\(\Rightarrow\left[{}\begin{matrix}2x+3\le5\\2x+3\ge-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x\le2\\2x\ge-8\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\le1\\x\ge-4\end{matrix}\right.\Rightarrow-4\le x\le1.\)
Vậy \(-4\le x\le1\)
