Question

Cho hpt :\(\int_{x+2y=3m+2}^{3x-y=2m-1}\)

Tìm m để hệ phương trình trên có nghiệm duy nhất (x;y) thoả mãn :

x² + y² =10

Answer 1

\(\left\{{}\begin{matrix}3x-y=2m-1\\x+2y=3m+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-2y=4m-2\\x+2y=3m+2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x=7m\\x+2y=3m+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m\\m+2y=3m+2\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=m\\2y=2m+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m\\y=m+1\end{matrix}\right.\)

x^2+y^2= 10

<=> m^2+(m+1)^2=10

<=> 2m^2+2m+1=10

<=> m= \(\dfrac{-1+\sqrt{19}}{2}\) hoặc m= \(\dfrac{-1-\sqrt{19}}{2}\)