Question

\(\left\{{}\begin{matrix}2x-y=m+1\\x+y=2m-1\end{matrix}\right.\)

Tìm m để hệ có nghiệm duy nhất (x;y) sao cho `x^2 -2y-1=0`.

Answer 1

\(\left\{{}\begin{matrix}2x-y=m+1\\x+y=2m-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x=3m\\2x-y=m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m\\y=m-1\end{matrix}\right.\)

Theo đề: \(x^2-2y-1=0\)

\(\Leftrightarrow m^2-2\left(m-1\right)-1=0\)

\(\Leftrightarrow m^2-2m+1=0\)

\(\Leftrightarrow\left(m-1\right)^2=0\Leftrightarrow m=1\).

Vậy: \(m=1.\)