\(\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{3}{y}=9\\\dfrac{7}{x}+\dfrac{4}{y}=17\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}4a+3b=9\\7a+4b=17\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}16a+12b=36\\21a+12b=51\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}-5a=-15\\4a+3b=9\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}a=3\\4.3+3b=9\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}a=3\\b=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=3\\\dfrac{1}{y}=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-1\end{matrix}\right.\)
Vậy ...
Ta có: \(\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{3}{y}=9\\\dfrac{7}{x}+\dfrac{4}{y}=17\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{28}{x}+\dfrac{21}{y}=63\\\dfrac{28}{x}+\dfrac{16}{y}=68\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y}=-5\\\dfrac{4}{x}+\dfrac{3}{y}=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{3}{-1}=9\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}-3=9\\y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}=12\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-1\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-1\end{matrix}\right.\)
