Ta có: \(6x^2+5y^2=74>6x^2\Leftrightarrow x^2< \dfrac{37}{3}\Leftrightarrow x^2\in\left\{0,1,4,9\right\}\)
\(x^2=0\Rightarrow x=0\) thay x=0 pt ta có:
\(6x^2+5y^2=74\\ \Leftrightarrow6.0^2+5y^2=74\\ \Leftrightarrow5y^2=74\\ \Leftrightarrow y^2=\dfrac{74}{5}\left(ktm\right)\)
\(x^2=1\Leftrightarrow x=\pm1\) thay x=\(\pm1\) pt ta có:
\(6x^2+5y^2=74\\ \Leftrightarrow6.\left(\pm1\right)^2+5y^2=74\\ \Leftrightarrow6+5y^2=74\\ \Leftrightarrow y^2=\dfrac{68}{5}\left(ktm\right)\)
\(x^2=4\Leftrightarrow x=\pm2\) thay x=\(\pm2\) pt ta có:
\(6x^2+5y^2=74\\ \Leftrightarrow6.\left(\pm2\right)^2+5y^2=74\\ \Leftrightarrow6.4+5y^2=74\\ \Leftrightarrow24+5y^2=74\\ \Leftrightarrow y^2=10\left(ktm\right)\)
\(x^2=9\Leftrightarrow x=\pm3\) thay x=\(\pm3\) vào pt ta có:
\(6x^2+5y^2=74\\ \Leftrightarrow6.\left(\pm3\right)^2+5y^2=74\\ \Leftrightarrow6.9+5y^2=74\\ \Leftrightarrow54+5y^2=74\\ \Leftrightarrow y^2=4\\ \Leftrightarrow y=\pm2\)
Vậy \(\left(x,y\right)\in\left\{\left(-3;-2\right);\left(-3;2\right);\left(3;-2\right);\left(3;2\right)\right\}\)
Ta có:
\(6\left(x^2-4\right)=5\left(10-y^2\right)\left(1\right)\)
\(\Rightarrow6\left(x^2-4\right)⋮5\Leftrightarrow\left(6;5\right)=1\)
\(\Rightarrow x^2-4⋮5\Leftrightarrow x^2=5k+4\left(k\inℕ\right)\)
Đặt \(\left(1\right)=x^2-4=5k\)ta lại có:
\(\Rightarrow y^2=10-6k\)
Mà \(\hept{\begin{cases}x^2>0\\y^2>0\end{cases}}\Rightarrow\hept{\begin{cases}5k+4>0\\10-6k>0\end{cases}}\)
\(\Rightarrow-\frac{4}{5}< k< \frac{5}{3}\Leftrightarrow\orbr{\begin{cases}k=0\left(loại\right)\\k=1\end{cases}}\)
\(k=1\Leftrightarrow\hept{\begin{cases}x^2=9\\y^2=4\end{cases}}\Rightarrow\hept{\begin{cases}x=\pm3\\y=\pm2\end{cases}}\)
Vậy cặp \(\left(x,y\right)\in\left\{\left(-3;-2\right);\left(3;2\right)\right\}\)
