Question

\(\left\{{}\begin{matrix}3x-y=2\\9x-my=m\end{matrix}\right.\)

tìm m để hệ pt có nghiệm x+y=2

Answer 1

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

Để HPT có nghiệm thì m ≠ 3

Có: x + y = 2

\(\Leftrightarrow\dfrac{-m+6}{m-3}+\dfrac{m}{3\left(m-3\right)}=2\)

\(\Leftrightarrow\dfrac{-3m+18+m}{3\left(m-3\right)}=2\)

\(\Leftrightarrow\dfrac{-2m+18}{3\left(m-3\right)}=2\)

\(\Leftrightarrow\dfrac{-m+9}{3\left(m-3\right)}=1\)

<=> -m + 9 = 3m - 9

<=> -4m + 18 = 0

\(\Leftrightarrow m=\dfrac{18}{4}\) (t/m)