Tìm a, b, c biết ab=c, bc=4a, ac=9b

a) Ta có:

\(\left\{{}\begin{matrix}ab=c\\bc=4a\\ac=9b\end{matrix}\right.\)

\(\Rightarrow\left(ab\right)\left(bc\right)\left(ac\right)=c\left(4a\right)\left(9b\right)\)

\(\Rightarrow\left(abc\right)^2=36\left(abc\right)\)

\(\Rightarrow\left(abc\right)^2-36\left(abc\right)=0\)

\(\Rightarrow abc\left(abc-36\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}abc=0\\abc=36\end{matrix}\right.\)

Nếu: \(abc=0\Rightarrow cc=0\Rightarrow c=0\Rightarrow4a=bc=0\Rightarrow\left\{{}\begin{matrix}a=0\\b=0\end{matrix}\right.\)

Nếu:

\(abc=36\Rightarrow\left(ab\right).c=cc=36\Rightarrow c=\pm6\)

\(c=6\Rightarrow4a=bc=6b\Rightarrow a=\dfrac{3b}{2}\)

Mà: \(ab=6\Rightarrow\dfrac{3b}{2}.b=6\Rightarrow b^2=6.\dfrac{2}{3}=4\Rightarrow b=\pm2\Rightarrow a=\pm3\)

Tương tự với \(c=-6\)

Vậy: \(\left(a;b;c\right)=\left(3;2;6\right);\left(-3;2;-6\right);\left(3;-2;-6\right);\left(-3;-2;6\right)\)

c) Ta có:

\(2\left(n-1\right)^2>0\forall n\)

\(\Rightarrow2\left(n-1\right)^2+3>3\forall n\)

\(\Rightarrow\dfrac{1}{2}\left(n-1\right)^2+3< \dfrac{1}{3}\forall n\)

Do đó \(Max_B=\dfrac{1}{3}\)

Dấu \("="\) xảy ra \(\Leftrightarrow2\left(n-1\right)^2=0\Leftrightarrow n=1\)

Vậy \(n=1\)