1) Ta có: \(x^2-3x-7\left(x-3\right)=0\)
\(\Leftrightarrow x\left(x-3\right)-7\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)
Vậy: S={3;7}
x2-3x-7(x-3)=0
x2-3x-7x+21=0
x2-10x+21=0
x=7 hoặc x=3
bài 1 :
Ta có: x2−3x−7(x−3)=0x2−3x−7(x−3)=0
⇔x(x−3)−7(x−3)=0
⇔x(x−3)−7(x−3)=0
⇔(x−3)(x−7)=0
⇔(x−3)(x−7)=0
⇔[x−3=0x−7=0
⇔[x=3x=7
⇔[x−3=0x−7=0
⇔[x=3x=7
Vậy: S={3;7}
