Question

[Lizy]

\(\left\{{}\begin{matrix}mx-y=2\\3x+my=5\end{matrix}\right.\)

tìm m để hệ có nghiệm (x;y) thỏa mãn `x+y=1 - (m^2)/(m^2 +3)`

Answer 1

[Nguyễn Việt Lâm]

\(\left\{{}\begin{matrix}mx-y=2\\3x+my=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m^2x-my=2m\\3x+my=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(m^2+3\right)x=2m+5\\y=mx-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m+5}{m^2+3}\\y=mx-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m+5}{m^2+3}\\y=\dfrac{5m-6}{m^2+3}\end{matrix}\right.\)

Thay vào \(x+y=1-\dfrac{m^2}{m^2+3}\)

\(\Leftrightarrow\dfrac{3m+5}{m^2+3}+\dfrac{5m-6}{m^2+3}=1-\dfrac{m^2}{m^2+3}\)

\(\Leftrightarrow\dfrac{8m-1}{m^2+3}=\dfrac{3}{m^2+3}\)

\(\Leftrightarrow8m-1=3\)

\(\Rightarrow m=\dfrac{1}{2}\)