a: Thay m=-3 vào hệ phương trình, ta được:
\(\left\{{}\begin{matrix}x-3y=2\\2x+4y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-6y=4\\2x+4y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-6y-2x-4y=4-3\\x-3y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-10y=1\\x=3y+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{10}\\x=3\cdot\dfrac{-1}{10}+2=2-\dfrac{3}{10}=\dfrac{17}{10}\end{matrix}\right.\)
b: Để hệ có nghiệm duy nhất thì \(\dfrac{1}{2}\ne\dfrac{m}{4}\)
=>\(m\ne\dfrac{4}{2}=2\)
\(\left\{{}\begin{matrix}x+my=2\\2x+4y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+2my=4\\2x+4y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x+2my-2x-4y=4-3\\x+2y=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y\left(2m-4\right)=1\\x=\dfrac{3}{2}-2y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{1}{2m-4}\\x=\dfrac{3}{2}-\dfrac{2}{2m-4}=\dfrac{3m-6-2}{2m-4}=\dfrac{3m-8}{2m-4}\end{matrix}\right.\)
x+y=0
=>\(\dfrac{3m-8+1}{2m-4}=0\)
=>3m-7=0
=>3m=7
=>\(m=\dfrac{7}{3}\left(nhận\right)\)
a) Với m = -3 ta có hpt:
\(\left\{{}\begin{matrix}x-3y=2\\2x+4y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-6y=4\\2x+4y=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-10y=1\\x-3y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{10}\\x+\dfrac{3}{10}=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{10}\\x=2-\dfrac{3}{10}=\dfrac{17}{10}\end{matrix}\right.\)
b) Để hpt có nghiệm duy nhất thì:
\(\dfrac{1}{2}\ne\dfrac{m}{4}\Leftrightarrow m\ne2\)
\(\left\{{}\begin{matrix}x+my=2\\2x+4y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+2my=4\\2x+4y=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(2m-4\right)y=1\\x+my=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{2m-4}\\x=2-my\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{2m-4}\\x=2-\dfrac{m}{2m-4}=\dfrac{4m-8-m}{2m-4}=\dfrac{3m-8}{2m-4}\end{matrix}\right.\)
Mà: x + y = 0 nên ta có:
\(\dfrac{3m-8}{2m-4}+\dfrac{1}{2m-4}=0\\ \Leftrightarrow\dfrac{3m-7}{2m-4}=0\\ \Leftrightarrow3m-7=0\Leftrightarrow3m=7\\ \Leftrightarrow m=\dfrac{7}{3}\left(tm\right)\)
