Có bao nhiêu cặp số nguyên \(\left( x\,;\,y \right)\) thỏa mãn

+ Điều kiện \({{3}^{x-2}}+2y>0\).

+ Phương trình tương đương: \({{2.3}^{x-1}}-{{\log }_{3}}\left( {{3}^{x-1}}+6y \right)=6y-x\,\,\,\left( * \right)\).

+ Đặt: \(u={{\log }_{3}}\left( {{3}^{x-1}}+6y \right)\Rightarrow {{3}^{x-1}}+6y={{3}^{u}}\) \(\Rightarrow 6y={{3}^{u}}-{{3}^{x-1}}\).

Ta có: \(\left( * \right)\Leftrightarrow {{2.3}^{x-1}}-u={{3}^{u}}-{{3}^{x-1}}-x\)

                 \(\Leftrightarrow {{3.3}^{x-1}}+x={{3}^{u}}+u\)\(\Leftrightarrow {{3}^{x}}+x={{3}^{u}}+u\).

+ Hàm \(f\left( t \right)={{3}^{t}}+t\) đồng biến trên \(\mathbb{R}\) nên

\({{3}^{x}}+x={{3}^{u}}+u\Leftrightarrow x=u\Leftrightarrow x={{\log }_{3}}\left( {{3}^{x-1}}+6y \right)\)

                       \(\Leftrightarrow {{3}^{x-1}}+6y={{3}^{x}}\)\(\Leftrightarrow y={{3}^{x-2}}\) (thỏa đk \({{3}^{x-2}}+2y>0\)).

+ Do \({{2022}^{-1}}\le y\le 2022\) nên \({{2022}^{-1}}\le {{3}^{x-2}}\le 2022\)

                     \(\Leftrightarrow {{\log }_{3}}{{2022}^{-1}}\le x-2\le {{\log }_{3}}2022\)

                     \(\Leftrightarrow {{\log }_{3}}{{2022}^{-1}}+2\le x\le {{\log }_{3}}2022+2\)

                     \(\Rightarrow -5<x<9\).

+ Do \(x\) nguyên, suy ra \(x\in \{-4;-3;....;8\}\).

                     \(x\in \{-4;-3;-2;-1;0;1\}\) suy ra \(y\) không nguyên do \(0<y={{3}^{x-2}}<1\).

                     \(x\in \{2;3;4;5;6;7;8\}\) suy ra \(y\) nguyên do \(y\in \{{{3}^{0}};{{3}^{1}};{{3}^{2}};{{3}^{3}};{{3}^{4}};{{3}^{5}};{{3}^{6}}\}\).

+ Vậy có 7 cặp số nguyên \(\left( x\,;\,y \right)\) thỏa YCBT.