Question

Cho pt:\(\left\{{}\begin{matrix}x+y=m\\2x-my=0\end{matrix}\right.\)(1)

Tìm m để hệ (1) có nghiệm (x;y) thỏa mãn :x+y=1

 

Answer 1

\(\left\{{}\begin{matrix}2x+2y=2m\\2x-my=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(m+2\right)y=2m\\x=m-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2m}{m+2}\\x=\dfrac{m^2+2m-2m}{m+2}=\dfrac{m^2}{m+2}\end{matrix}\right.\)

Thay vào ta được 

\(\dfrac{m^2+2}{m+2}=1\Leftrightarrow m^2+2=m+2\Leftrightarrow m^2-m=0\Leftrightarrow m=0;m=1\)