Câu 1
PT <=> 3x(x-2) - 7(x-2) = 0
<=> (x-2)(3x-7) = 0
<=> \(\left[{}\begin{matrix}x-2=0< =>x=2\\3x-7=0< =>x=\frac{7}{3}\end{matrix}\right.\)
KL: ...
Câu 2:
\(\left(x-3\right)^2=4\)
<=> \(\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
=> Chọn D