Question

tìm m để hệ phương trình \(\left\{{}\begin{matrix}\frac{2}{x}+\frac{6}{y}=11\\\frac{4}{x}-\frac{9}{y}=1\end{matrix}\right. \) có nghiệm duy nhất

Answer 1

ĐKXĐ : \(x;y\ne0\)

\(\left\{{}\begin{matrix}\frac{2}{x}+\frac{6}{y}=11\\\frac{4}{x}-\frac{9}{y}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{4}{x}+\frac{12}{y}=22\\\frac{4}{x}-\frac{9}{y}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{21}{y}=21\\\frac{4}{x}-\frac{9}{y}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=\frac{2}{5}\end{matrix}\right.\)

Vậy...

Answer 2

Cách khác :

\(\left\{{}\begin{matrix}\frac{2}{x}+\frac{6}{y}=11\\\frac{4}{x}-\frac{9}{y}=1\end{matrix}\right.\)

ĐKXĐ : \(x\ne0\); \(y\ne0\)

Đặt \(\left\{{}\begin{matrix}a=\frac{1}{x}\\b=\frac{1}{y}\end{matrix}\right.\) (*)

Thay vào HPT ta có : \(\left\{{}\begin{matrix}2a+6b=11\\4a-9b=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4a+12b=22\\4a-9b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}21b=21\\2a+6b=11\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=1\\2a+6.1=11\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=1\\a=\frac{5}{2}\end{matrix}\right.\)

Thay vào (*) ta có \(\left\{{}\begin{matrix}a=\frac{1}{x}=\frac{5}{2}\Rightarrow x=\frac{2}{5}\\b=1\Rightarrow y=1\end{matrix}\right.\)

Vậy....