\(x^2-x=0\\ \Leftrightarrow x\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
15,\(x^2-5x+6=x^2-2x-3x+6=\left(x^2-2x\right)-\left(3x-6\right)=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
x2-x=0
-2x2=0
=>-2x2=0 nếu x=0
Vậy x=0
Câu 15: x2-5x+6=x2-3x-2x+6=(x2-3x)-(2x-6)=x(x-3)-2(x-3)
=(x-2)(x-3)