Điều kiện xác định \(x\le3\)
\(\left\{{}\begin{matrix}\sqrt[]{3-x}=a\\\sqrt{4-x}=b\\\sqrt{5-x}=c\end{matrix}\right.\left(a,b,c\ge0\right)\Leftrightarrow\left\{{}\begin{matrix}x=3-a^2\\x=4-b^2\\x=5-c^2\end{matrix}\right.\)
\(PT\Leftrightarrow x=ab+bc+ca\)
\(\Leftrightarrow\left\{{}\begin{matrix}ab+bc+ca=3-a^2\\ab+bc+ca=4-b^2\\ab+bc+ca=5-c^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}ab+bc+ca+a^2=3\\ab+bc+ca+b^2=4\\ab+bc+ca+c^2=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)\left(a+c\right)=3\left(1\right)\\\left(b+c\right)\left(b+a\right)=4\left(2\right)\\\left(c+a\right)\left(c+b\right)=5\left(3\right)\end{matrix}\right.\)
\(\Rightarrow\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2=60\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=2\sqrt{15}\left(4\right)\)
\(\left(4\right):\left(1\right),\left(2\right),\left(3\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=\dfrac{2\sqrt[]{15}}{5}\\a+c=\dfrac{\sqrt[]{15}}{2}\\b+c=\dfrac{2\sqrt[]{15}}{3}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=\dfrac{2\sqrt{15}}{5}\\a-b=-\dfrac{\sqrt{15}}{6}\end{matrix}\right.\)
\(\Rightarrow a=\dfrac{7\sqrt{15}}{30}\Rightarrow x=3-a^2=\dfrac{131}{60}\)
\(\Rightarrow S=\left\{\dfrac{131}{60}\right\}\)
