Question

tim m de hpt \(\hept{\begin{cases}mx+4y=10-m\\x+my=4\end{cases}}\) có nghiệm duy nhất (x;y) sao cho x>0;y>0

Answer 1

\(\hept{\begin{cases}mx+4y=10-m\\x+my=4\end{cases}}\)

\(\hept{\begin{cases}m.\left(4-my\right)+m=10\\x=4-my\end{cases}}\)

\(\hept{\begin{cases}m\left(5-my\right)=10\\x=4-my\end{cases}}\)

\(\hept{\begin{cases}x=4-my\\m=\frac{10}{5-my}\end{cases}}\)

\(\hept{\begin{cases}x=4-\frac{10y}{5-my}\\4-\frac{10y}{5-my}=\frac{10}{5-my}\end{cases}}\)

\(\hept{\begin{cases}x=4-\frac{10y}{5-my}\\\frac{-10y-10}{5-my}=-4\end{cases}}\)