x2-4x+3=0
⇔ x\(^2\)-x-3x+3=0
⇔ x(x-1)-3(x-1)=0
⇔( x-3)(x-1)=0
⇔\(\left[{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy x ∈\(\left\{3;1\right\}\)
<=> x2-2.2.x+22-1=0
<=> (x-2)2-1=0
<=> (x-2-1)(x-2+1)=0
<=> (x-3)(x-1)=0
<=> \(\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy x=3 hoặc x=-1
\(x^2\) -4x+3=0
=> \(x^2\)-x-3x+3=0
=> x.(x-1)+3.(x-1)
=> (x-3).(x-1)
\(\begin{cases} x-3=0 \\ x-1=0 \end{cases}\)=> \(\begin{cases} x=3 \\ x=1 \end{cases}\)
Vậy x=3
x=1