(x-1)(x-5)+3=0
=>\(^{x^2}\)- 5x - x + 5 + 3 = 0
=>\(x^2\) - 6x + 8 =0
=>( \(x^2\) - 6x + 9 ) - 1 = 0
=>\(\left(x-3\right)^2\) - 1 = 0
=> ( x - 3 - 1 )( x - 3 + 1 ) = 0
=> ( x - 4 )( x - 2 ) = 0
=> * x - 4 = 0
x = 4
* x - 2 = 0
x = 2
Vậy x = 4; 2