\(\left\{{}\begin{matrix}2x+y=3m-1\\x-2y=-m-3\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{3m-1-y}{2}\\\dfrac{3m-1-y}{2}-2y=-m-3\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{3m-1-y}{2}\\3m-1-y-4y=-2m-6\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{3m-1-y}{2}\\5y=5m+5\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{3m-1-y}{2}\\y=m+1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{3m-1-m-1}{2}\\y=m+1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=m-1\\y=m+1\end{matrix}\right.\)
Vậy hpt trên có nghiệm duy nhất \(\left\{{}\begin{matrix}x=m-1\\y=m+1\end{matrix}\right.\)
Ta có: y = x2 \(\Leftrightarrow\) m + 1 = (m - 1)2 \(\Leftrightarrow\) m + 1 = m2 - 2m + 1
\(\Leftrightarrow\) m2 - 3m = 0
\(\Leftrightarrow\) m(m - 3) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}m=0\\m-3=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}m=0\\m=3\end{matrix}\right.\)
Vậy m = 0; m = 3 thì hpt trên có nghiệm duy nhất và thỏa mãn y = x2
Chúc bn học tốt!
