c) |-x+7|=24
⇒\(\left[{}\begin{matrix}-x+7=24\\-x+7=-24\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-17\\x=31\end{matrix}\right.\)
d) |x+8|+15=0
|x+8|=0-15
|x+8|=-15
⇒x=∅
e) |x|+|x-3|=0
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
c: Ta có: \(\left|-x+7\right|=24\)
\(\Leftrightarrow\left[{}\begin{matrix}7-x=24\\7-x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-17\\x=31\end{matrix}\right.\)