Question

Cho hệ phương trình: \(\left\{{}\begin{matrix}3x-my=2\\9x-6y=1\end{matrix}\right.\)

Tìm m để hệ phương trình có nghiệm duy nhất (x;y) thỏa mãn: \(\left\{{}\begin{matrix}x>0\\y< 0\end{matrix}\right.\)

Answer 1

Để hệ có nghiệm duy nhất thì \(\dfrac{3}{9}\ne\dfrac{-m}{-6}\)

=>\(\dfrac{m}{6}\ne\dfrac{1}{3}\)

=>\(m\ne2\)

\(\left\{{}\begin{matrix}3x-my=2\\9x-6y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x-3my=6\\9x-6y=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}9x-3my-9x+6y=6-1\\9x-6y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y\left(-3m+6\right)=5\\9x=6y+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=\dfrac{5}{-3m+6}\\x=\dfrac{6y+1}{9}=\dfrac{6\cdot\dfrac{5}{-3m+6}+1}{9}=\dfrac{\dfrac{30-3m+6}{-3m+6}}{9}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=\dfrac{5}{-3m+6}\\x=\dfrac{\dfrac{-3m+36}{-3m+6}}{9}=\dfrac{m-12}{m-2}:9=\dfrac{m-12}{9\left(m-2\right)}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x>0\\y< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{m-12}{9\left(m-2\right)}>0\\\dfrac{5}{-3m+6}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{m-12}{m-2}>0\\-3m+6< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}m>12\\m< 2\end{matrix}\right.\\-3m< -6\end{matrix}\right.\Leftrightarrow m>12\)

Answer 2

Để hpt có nghiệm duy nhất thì:

\(\dfrac{3}{9}\ne\dfrac{-m}{-6}\Leftrightarrow m\ne2\)

\(\left\{{}\begin{matrix}3x-my=2\\9x-6y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x-3my=6\\9x-6y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(6-3m\right)y=5\\9x-6y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5}{6-3m}\\9x-\dfrac{30}{6-3m}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5}{6-3m}\\9x=1+\dfrac{30}{6-3m}=\dfrac{36-3m}{6-3m}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5}{6-3m}\\x=\dfrac{12-m}{2-m}:9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5}{6-3m}\\x=\dfrac{12-m}{9\left(2-m\right)}\end{matrix}\right.\)

Mà: \(\left\{{}\begin{matrix}x>0\\y< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12-m}{9\left(2-m\right)}>0\\\dfrac{5}{6-3m}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m>12\\m< 2\end{matrix}\right.\\m>2\end{matrix}\right.\Leftrightarrow m>12\)