\({{{x^{n + 2}}{y^{3 – m}}} \over {{x^m}{y^{2n + 5}}}} = {{{x^{13}}} \over {{y^9}}}\)
\(\eqalign{ & \,\,\,\,\,\,\,{{{x^{n + 2}}{y^{3 – m}}} \over {{x^m}{y^{2n + 5}}}} = {{{x^{13}}} \over {{y^9}}} \cr & \Leftrightarrow {x^m}{y^{2n + 5}}{x^{13}} = {x^{n + 2}}{y^{3 – m}}{y^9} \cr & \Leftrightarrow {x^m}{x^{13}}{y^{2n + 5}} = {x^{n + 2}}{y^{3 – m + 9}} \cr & \Leftrightarrow {x^{m + 13}}{y^{2n + 5}} = {x^{n + 2}}{y^{12 – m}} \cr & \Leftrightarrow \left\{ \matrix{ m + 13 = n + 2 \hfill \cr 2n + 5 = 12 – m \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ m = n – 11 \hfill \cr 2n + m = 7 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ m = n – 11 \hfill \cr 2n + n – 11 = 7 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ m = n – 11 \hfill \cr 3n = 18 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ m = 6 – 11 \hfill \cr n = 6 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ m = – 5 \hfill \cr n = 6 \hfill \cr} \right. \cr} \)
Vậy \(m = – 5\) và \(n = 6\) thì \({{{x^{n + 2}}{y^{3 – m}}} \over {{x^m}{y^{2n + 5}}}} = {{{x^{13}}} \over {{y^9}}}\)