\(\left\{{}\begin{matrix}\frac{14}{x-y+2}-\frac{10}{x+y-1}=9\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}a=\frac{1}{x-y+2}\\b=\frac{1}{x+y-1}\end{matrix}\right.\)
Thay a,b vào hệ phương trình ta có:
\(\left\{{}\begin{matrix}14a-10b=9\\3a+2b=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}14a-10b=9\\15a+10b=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}29a=29\\3a+2b=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\3.1+2b=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=\frac{1}{2}\end{matrix}\right.\)
Ta có:
\(a=\frac{1}{x-y+2}=1\Leftrightarrow x-y+2=1\)
\(b=\frac{1}{x+y-1}=\frac{1}{2}\Leftrightarrow x+y-1=2\)
Ta có hệ phương trình mới:
\(\left\{{}\begin{matrix}x-y+2=1\\x+y-1=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\x+y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=2\\x+y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\1+y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
đk: x-y+2\(\ne0;x+y-1\ne0\)
\(\left\{{}\begin{matrix}\frac{14}{x-y+2}-\frac{10}{x+y-1}=9\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{14}{x-y+2}-\frac{10}{x+y-1}=9\\\frac{15}{x-y+2}+\frac{10}{x+y-1}=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{29}{x-y+2}=29\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-y+2=1\\\frac{2}{x+y-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\x+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{3}{2}\end{matrix}\right.\)(tm)
vậy hệ phương trình có nghiệm (x;y)=(1/2;3/2)
