ĐKXĐ : \(\left\{{}\begin{matrix}x\ne1\\y\ne2\end{matrix}\right.\)
Xét hệ phương trình :
\(\left\{{}\begin{matrix}\frac{2}{x-1}+\frac{3}{y-2}=\frac{13}{11}\\\frac{3}{x-1}-\frac{4}{y-2}=-3\end{matrix}\right.\)
Đặt : \(\left\{{}\begin{matrix}a=\frac{1}{x-1}\\b=\frac{1}{y-2}\end{matrix}\right.\) \(\left(1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}2a+3b=\frac{11}{3}\\3a-4b=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6a+9b=11\\6a-8b=-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}17b=17\\3a-4b=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=1\\a=\frac{1}{3}\end{matrix}\right.\)
Thay vào \(\left(1\right)\) ta có :
\(\left\{{}\begin{matrix}\frac{1}{3}=\frac{1}{x-1}\\1=\frac{1}{y-2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=3\end{matrix}\right.\)
Vậy..
Điều kiện: \(x\ne1\) và \(y\ne2\)
Đặt \(\frac{1}{x-1}=a\) và \(\frac{1}{y-2}=b\)
\(\Rightarrow\left\{{}\begin{matrix}2a+3b=\frac{11}{3}\\3a-4b=-3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6a+9b=11\\6a-8b=-6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}17b=17\\a=\frac{4b-3}{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=1\\a=\frac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x-1}=\frac{1}{3}\\\frac{1}{y-2}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=3\end{matrix}\right.\) (Thỏa mãn)
Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(4;3\right)\)
