Question

(x-1)(x-5)+3=0

Answer 1

\(\Leftrightarrow x^2-5x-x+5+3=0\)

\(\Leftrightarrow x^2-6x+8=0\)

\(\Leftrightarrow x^2-4x-2x+8=0\)

\(\Leftrightarrow x.\left(x-4\right)-2.\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-2\right).\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

Answer 2

( x - 1 )( x - 5 ) + 3 = 0

( x² - 5x - x +5 ) = 3

( x² - 6x + 5 ) = 3

( x² - x - 5x + 5 ) = 3

[ x( x - 1 ) - 5( x - 1 ) ] = 3

( x - 5 )( x - 1 ) = 3

-> TH1: x - 5 = 3 -> x = 8

TH2 : x - 1 = 3 x = 4

Vậy x thuộc { 8 ; 4 }