1. ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\frac{x+3}{x-2}=a\\\frac{x-3}{x+2}=b\end{matrix}\right.\) ta được:
\(a^2+6b^2=7ab\Leftrightarrow a^2-7ab+6b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(a-6b\right)=0\Rightarrow\left[{}\begin{matrix}a=b\\a=6b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x+3}{x-2}=\frac{x-3}{x+2}\\\frac{x+3}{x-2}=\frac{6x-18}{x+2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)\left(x+2\right)=\left(x-3\right)\left(x-2\right)\\\left(x+3\right)\left(x+2\right)=\left(6x-18\right)\left(x-2\right)\end{matrix}\right.\)
Chắc bạn tự làm đoạn còn lại được
2.
\(x^2+y^2-2xy+5\left(x^2y^2-7xy+12\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+5\left(xy-3\right)\left(xy-4\right)=0\)
Do \(\left(x-y\right)^2\ge0;\forall x;y\Rightarrow5\left(xy-3\right)\left(xy-4\right)\le0\)
\(\Leftrightarrow3\le xy\le4\)
Mà x;y nguyên nên dấu "=" xảy ra khi và chỉ khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\xy=3\end{matrix}\right.\\\left\{{}\begin{matrix}x=y\\xy=4\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(2;2\right);\left(-2;-2\right)\)
