N.(N + 1) = 12
N2 + N = 12
N2 + N - 12 = 0
(N - 3)(N - 4) = 0
N - 3 = 0 hoặc N - 4 = 0
N = 3 hoặc N = 4
\(N.\left(N+1\right)=12\)
\(N^2+N=12\)
\(N^2+N-12=0\)
\(N^2+4N-3N-12=0\)
\(N\left(N+4\right)-3\left(N+4\right)=0\)
\(\left(N+4\right)\left(N-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}N+4=0\\N-3=0\end{cases}\Rightarrow\orbr{\begin{cases}N=-4\\N=3\end{cases}}}\)
VẬY N=-4 ; N=3
\(n\times\left(n+1\right)=12\)
\(n^2+n=12\)
\(n^2+n-12=12-\)\(12\)
\(n^2+n-12=0\)
\(n^2-3n+4n-12=\)\(0\)
\(n\times\left(\frac{n^2}{n}-\frac{3n}{n}\right)\)\(+2^2\times\)\(\left(\frac{2^2\times n}{2^2}-\frac{2^2\times3}{2^2}\right)=\)\(0\)
\(n\times\left(n-3\right)+2^2\times\left(n-3\right)=\)\(0\)
\(\left(n-3\right)\times\)\(\left(\frac{n\times\left(n-3\right)}{n-3}+\frac{2^2\times\left(n-3\right)}{n-3}\right)\)\(=0\)
\(\left(n-3\right)\times\left(n+4\right)=0.\)
\(\hept{\begin{cases}n-3=0\\n+4=0\end{cases}}\)
\(\hept{\begin{cases}\left(n-3\right)+3=3\\\left(n+4\right)+\left(-4\right)=-4\end{cases}}\)
\(\hept{\begin{cases}n-3+3=3\\\left(n+4\right)-4=-4\end{cases}}\)
\(\hept{\begin{cases}n=3\\n=-4\end{cases}}\)
\(n\in\left\{3,-4\right\}\)
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