Bài 7:
\(a,m=1\Leftrightarrow\left\{{}\begin{matrix}x-y=1\\x+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=3\end{matrix}\right.\\ b,HPT\Leftrightarrow\left\{{}\begin{matrix}y=1-mx\left(1\right)\\x+m\left(1-mx\right)=m+6\left(2\right)\end{matrix}\right.\\ \left(2\right)\Leftrightarrow x+m-m^2x=m+6\\ \Leftrightarrow x\left(1-m^2\right)=6\Leftrightarrow x=\dfrac{6}{1-m^2}\)
Thay vào \(\left(1\right)\Leftrightarrow y=1-\dfrac{6m}{1-m^2}=\dfrac{1-m^2-6m}{1-m^2}\)
\(3x-y=1\Leftrightarrow\dfrac{18}{1-m^2}-\dfrac{1-m^2-6m}{1-m^2}=1\\ \Leftrightarrow\dfrac{m^2+6m-17}{1-m^2}=1\\ \Leftrightarrow m^2+6m-17=1-m^2\\ \Leftrightarrow2m^2+6m-18=0\\ \Leftrightarrow m^2+3m-9=0\\ \Leftrightarrow\left[{}\begin{matrix}m=\dfrac{3\sqrt{5}-3}{2}\\m=\dfrac{-3\sqrt{5}-3}{2}\end{matrix}\right.\)
