Chọn B
Ta có
\(\[P=x+2y+\frac{5x+5y}{xy-1} \] \)
\(\[=\frac{5\left(x+y\right)}{xy-1} +\frac{5\left(xy-1\right)}{x+y} +x+2y-\frac{5xy-5}{x+y} \] \)
\(\[\ge 2.5+\frac{x^{2} +3xy+2y^{2} -5xy+5}{x+y} \] \)
\(\[=2.5+\frac{x^{2} -2xy+2y^{2} +5}{x+y} \] \)
\(\[=12+\frac{x^{2} -2xy+2y^{2} -2x-2y+5}{x+y} \] \)
\(\[=12+\frac{\left(2x-3y\right)^{2} +3\left(y-2\right)^{2} +2\left(x-3\right)^{2} }{6\left(x+y\right)} \ge 12.\] \)
Đẳng thức xảy ra khi \(\left\{\begin{array}{l} {2x-3y=0} \\ {x-3=0} \\ {y-2=0} \\ {\frac{5\left(x+y\right)}{xy-1} =\frac{5\left(xy-1\right)}{x+y} } \end{array}\right. \Leftrightarrow \left\{\begin{array}{l} {x=3} \\ {y=2} \end{array}\right. . \)
Do đó\( S=\frac{x_{0} +1}{y_{0} } =\frac{3+1}{2} =2. \)