Vì \(\dfrac{1}{2}\ne\dfrac{2}{5}\)
nên hệ luôn có nghiệm duy nhất
\(\left\{{}\begin{matrix}x+2y=m\\2x+5y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+4y=2m\\2x+5y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x+5y-2x-4y=1-2m\\x+2y=m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2m+1\\x=m-2\left(-2m+1\right)=m+4m-2=5m-2\end{matrix}\right.\)
|x|=y
=>\(-2m+1=\left|5m-2\right|\)
=>\(\left\{{}\begin{matrix}-2m+1>=0\\\left(5m-2\right)^2=\left(-2m+1\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< =\dfrac{1}{2}\\\left(5m-2+2m-1\right)\left(5m-2-2m+1\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< =\dfrac{1}{2}\\\left(7m-3\right)\left(3m-1\right)=0\end{matrix}\right.\Leftrightarrow m\in\left\{\dfrac{3}{7};\dfrac{1}{3}\right\}\)
Ta có:
\(\left\{{}\begin{matrix}x+2y=m\\2x+5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+4y=2m\\2x+5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1-2m\\x+2y=m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1-2m\\x+2-4m=m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1-2m\\x=5m-2\end{matrix}\right.\)
Vậy hệ pt luôn có 2 nghiệm \(\left(x;y\right)\) là:\(\left(5m-2;1-2m\right)\)
Để \(\left|x\right|=y\)
Thì \(\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5m-2=1-2m\\5m-2=2m-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}7m=3\\3m=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{3}{7}\\m=\dfrac{1}{3}\end{matrix}\right.\)
