Giải hpt:
\(\left\{{}\begin{matrix}3\left(a+\frac{1}{a}\right)=4\left(b+\frac{1}{b}\right)=5\left(c+\frac{1}{c}\right)\\ab+bc+ca=1\end{matrix}\right.\)
\(\frac{3\left(a^2+1\right)}{a}=\frac{4\left(b^2+1\right)}{b}=\frac{5\left(c^2+1\right)}{c}\)
\(\Leftrightarrow\frac{3\left(a+b\right)\left(a+c\right)}{a}=\frac{4\left(a+b\right)\left(b+c\right)}{b}=\frac{5\left(a+c\right)\left(b+c\right)}{c}\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{3\left(a+c\right)}{a}=\frac{4\left(b+c\right)}{b}\\\frac{4\left(a+b\right)}{b}=\frac{5\left(a+c\right)}{c}\\\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3ab+3bc=4ab+4ac\\4ac+4bc=5ab+5bc\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}ab-3bc+4ac=0\\5ab+bc-4ac=0\end{matrix}\right.\)
Đặt \(\left(ab;bc;ca\right)=\left(x;y;z\right)\) và kết hợp \(ab+bc+ca=1\) ta có hệ:
\(\left\{{}\begin{matrix}x+y+z=1\\x-3y+4z=0\\5x+y-4z=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{6}\\y=\frac{1}{2}\\z=\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}ab=\frac{1}{6}\\bc=\frac{1}{2}\\ac=\frac{1}{3}\end{matrix}\right.\) (1)
\(\Rightarrow\left(abc\right)^2=\frac{1}{36}\Rightarrow abc=\pm\frac{1}{6}\) (2)
Chia vế cho vế (2) cho (1) sẽ tìm nốt được a;b;c
