Vì \(\dfrac{2}{1}\ne\dfrac{1}{-2}\)
nên hệ luôn có nghiệm duy nhất
\(\left\{{}\begin{matrix}2x+y=5m-1\\x-2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y=5m-1\\2x-4y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x+y-2x+4y=5m-1-4\\x-2y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5y=5m-5\\x=2y+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=m-1\\x=2\left(m-1\right)+2=2m-2+2=2m\end{matrix}\right.\)
\(x^2+2y^2=2\)
=>\(\left(2m\right)^2+2\cdot\left(m-1\right)^2=2\)
=>\(4m^2+2m^2-4m+2-2=0\)
=>\(6m^2-4m=0\)
=>m(6m-4)=0
=>\(\left[{}\begin{matrix}m=0\\6m-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=0\\m=\dfrac{2}{3}\end{matrix}\right.\)
Ta có:\(2\ne-\dfrac{1}{2}\)
\(\Rightarrow\)Hệ pt luôn có nghiệm duy nhất với mọi \(m\)
Lại có:
\(\left\{{}\begin{matrix}2x+y=5m-1\\x-2y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5m-1\\2x-4y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5y=5m-5\\x-2y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5m-5}{5}\\x=2+2y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=m-1\\x=2+2m-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=m-1\\x=2m\end{matrix}\right.\)
Theo đề bài ta có:
\(x^2+2y^2=2\)
\(\Leftrightarrow\left(2m\right)^2+2\left(m-1\right)^2=2\)
\(\Leftrightarrow4m^2+2\left(m^2-2m+1\right)=2\)
\(\Leftrightarrow4m^2+2m^2-4m+2=2\)
\(\Leftrightarrow6m^2-4m=0\)
\(\Leftrightarrow2m\left(3m-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2m=0\\3m-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}m=0\\m=\dfrac{2}{3}\end{matrix}\right.\)
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