Câu 1:
a: 6x-24=0
=>6x=24
hay x=4
b: \(\left|-3x-9\right|=5x+3\)
\(\Leftrightarrow\left|3x+9\right|=5x+3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{3}{5}\\\left(5x+3-3x-9\right)\left(5x+3+3x+9\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{3}{5}\\\left(2x-6\right)\left(8x+12\right)=0\end{matrix}\right.\Leftrightarrow x=3\)