`a, (21 x -3)(x-2)=0`
`-> 3(7x-1)(x-2) = 0`
`-> x = 1/7` hoặc `2`
`b, -(x^2 - 4x + 3) = 0`
`-> -(x-1)(x-3) =0`
`->` \(\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\)
`->` \(\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
b) -x² + 4x - 3 = 0
-x² + x + 3x - 3 = 0
-x(x - 1) + 3(x - 1) = 0
(x - 1)(-x + 3) = 0
x - 1 = 0 hoặc -x + 3 = 0
*) x - 1 = 0
x = 0 + 1
x = 1
*) -x + 3 = 0
-x = 0 - 3
-x = -3
x = 3
Vậy x = 1; x = 3