ĐKXĐ: \(x\le3\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{3-x}\\b=\sqrt{4-x}\\c=\sqrt{5-x}\end{matrix}\right.\) \(a;b;c\ge0\) \(\Rightarrow\left\{{}\begin{matrix}x=3-a^2\\x=4-b^2\\x=5-c^2\end{matrix}\right.\) (1)
Từ pt ban đầu ta có: \(x=ab+ac+bc\)
Thế vào (1): \(\left\{{}\begin{matrix}ab+ac+bc=3-a^2\\ab+ac+bc=4-b^2\\ab+ac+bc=5-c^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2+ab+ac+bc=3\\b^2+ab+ac+bc=4\\c^2+ab+ac+bc=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)\left(a+c\right)=3\\\left(a+b\right)\left(b+c\right)=4\\\left(a+c\right)\left(b+c\right)=5\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}a+b=X\\a+c=Y\\b+c=Z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}XY=3\\XZ=4\\YZ=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\frac{Y}{Z}=\frac{3}{4}\\YZ=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=\frac{2\sqrt{15}}{5}\\a+c=\frac{\sqrt{15}}{2}\\b+c=\frac{2\sqrt{15}}{3}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=\frac{2\sqrt{15}}{5}\\a-b=\frac{-\sqrt{15}}{6}\end{matrix}\right.\) \(\Rightarrow a=\frac{7\sqrt{15}}{30}\) \(\Rightarrow x=3-a^2=\frac{131}{60}\) (t/m)
