Bài 6:
a: Thay m=2 vào hệ phương trình, ta được:
\(\left\{{}\begin{matrix}x-2y=5\\2x-y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-4y=10\\2x-y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y-2x+y=10-4\\2x-y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-3y=6\\2x-y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\2x=y+4=-2+4=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
b: Để hệ có nghiệm duy nhất thì \(\dfrac{1}{m}< >-\dfrac{2}{-1}=2\)
=>\(m\ne\dfrac{1}{2}\)
\(\left\{{}\begin{matrix}x-2y=5\\mx-y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=mx-4\\x-2\left(mx-4\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=mx-4\\x\left(1-2m\right)+8=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{-3}{1-2m}=\dfrac{3}{2m-1}\\y=m\cdot\dfrac{3}{2m-1}-4=\dfrac{3m-4\left(2m-1\right)}{2m-1}=\dfrac{-5m+4}{2m-1}\end{matrix}\right.\)
Để x và y trái dấu thì x*y<0
=>\(\dfrac{3\left(-5m+4\right)}{\left(2m-1\right)^2}< 0\)
=>-5m+4<0
=>-5m<-4
=>\(m>\dfrac{4}{5}\)
c: x=|y|
=>\(\dfrac{3}{2m-1}=\left|\dfrac{-5m+4}{2m-1}\right|\)
=>\(\left[{}\begin{matrix}\dfrac{-5m+4}{2m-1}=\dfrac{3}{2m-1}\\\dfrac{-5m+4}{2m-1}=\dfrac{-3}{2m-1}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-5m+4=3\\-5m+4=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}-5m=-1\\-5m=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{1}{5}\left(nhận\right)\\m=\dfrac{7}{5}\left(loại\right)\end{matrix}\right.\)
