a/ \(3x-5=13\\↔3x=18\\↔x=6\)
Vậy \(x=6\)
b/ \( (x-2)(x-3)=0\\↔\left[\begin{array}{1}x-2=0\\x-3=0\end{array}\right.\\↔\left[\begin{array}{1}x=2\\x=3\end{array}\right.\)
Vậy \(x=2\) hoặc \(x=3\)
c/ \( |x-2|=4\\↔\left[\begin{array}{1}x-2=4\\x-2=-4\end{array}\right.\\↔\left[\begin{array}{1}x=6\\x=-2\end{array}\right.\)
Vậy \(x=6\) hoặc \(x=-2\)
d/ \(3x-2\vdots x-1\\↔3x-3+1\vdots x-1\\↔3(x-1)+1\vdots x-1\)
Vì \( 3(x-1)\vdots x-1\) mà \( 3(x-1)+1\vdots x-1\)
\(→1\vdots x-1\)
\(→x-1∈Ư(1)=\{±1\}\)
\(→x∈\{2;0\}\)
Vậy \(x∈\{2;0\}\)
