$n^2-3n-18=0$ $⇔(n^2-6n)+(3n-18)=0$ $⇔n.(n-6)+3.(n-6)=0$$⇔(n-6)(n+3)=0$ $⇔\left[ \begin{array}{l}n-6=0\\n+3=0\end{array} \right.⇔\left[ \begin{array}{l}n=6\\n=-3\end{array} \right.$ Vậy $S=\{6;-3\}$. Bình luận
n²-3n-18=0 ⇔ n²+3n-6n-18=0 ⇔ n(n+3)-6(n+3)=0 ⇔ (n+3)(n-6)=0 ⇔ \(\left[ \begin{array}{l}n=-3 \\n=6\end{array} \right.\) Bình luận
$n^2-3n-18=0$
$⇔(n^2-6n)+(3n-18)=0$
$⇔n.(n-6)+3.(n-6)=0$
$⇔(n-6)(n+3)=0$
$⇔\left[ \begin{array}{l}n-6=0\\n+3=0\end{array} \right.⇔\left[ \begin{array}{l}n=6\\n=-3\end{array} \right.$
Vậy $S=\{6;-3\}$.
n²-3n-18=0
⇔ n²+3n-6n-18=0
⇔ n(n+3)-6(n+3)=0
⇔ (n+3)(n-6)=0
⇔ \(\left[ \begin{array}{l}n=-3 \\n=6\end{array} \right.\)