a) Ta có: 3x-17=x+3
\(\Leftrightarrow3x-x=3+17\)
\(\Leftrightarrow2x=20\)
hay x=10
b) Ta có: \( \left|x-3\right|-12=\left|-5\right|\)
\(\Leftrightarrow\left|x-3\right|=17\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=17\\x-3=-17\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=20\\x=-14\end{matrix}\right.\)
Vậy: \(x\in\left\{20;-14\right\}\)
c) Ta có: \(25-\left(x-5\right)=-415-\left(15-415\right)\)
\(\Leftrightarrow25-x+5=-415-15+415\)
\(\Leftrightarrow30-x=-15\)
hay x=45
Vậy: x=45
3x - 17 = x + 3
3x = x+3 +17
3x = x+20
3x-x=20
2x = 20
x= 20:2
x = 10
b ) |x-3| -12 = |-5|
|x-3| = 12 + 5
|x-3|=17
\(\left[{}\begin{matrix}x-3=17\\x-3=-17\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=17-3\\x=-17-3\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=14\\x=-20\end{matrix}\right.\)
vậy x= 14 hoặc x= -20
c) 25 -(x-5) = -415 -(15-145)
25-(x-5) = 15
x-5 = 25 -15
x-5 = 10
x = 10 - 5
x = 5
