a)
x(x - 2) + x - 2 = 0
Tương đương (x - 2)(x + 1) = 0
Hoặc x - 2 = 0 => x = 2
Hoặc x + 1 = 0 => x = -1
Vậy x = -1; x = 2
b)
5x(x - 3) - x + 3 = 0
<=> 5x(x - 3) - (x - 3) = 0
<=> (x - 3)(5x - 1) = 0
Hoặc x - 3 = 0 => x = 3
Hoặc 5x - 1 = 0 => x = 1/5.
Vậy x = 1/5; x = 3.
Câu a :
\(x\left(x-2\right)-x+2=x\left(x-2\right)-\left(x-2\right)=\left(x-2\right)\left(x-1\right)\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\Rightarrow x=2\\x-1=0\Rightarrow x=1\end{matrix}\right.\)
Câu b : Sửa đề
\(x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\Rightarrow x=2\\x-3=0\Rightarrow x=3\end{matrix}\right.\)
