\(\left\{{}\begin{matrix}3x+2y=5\\4x-3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12x+8y=20\\12x-9y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)
Thay x=1 và y=1 vào (d), ta được:
3m-1+1=m-2
=>3m=m-2
=>2m=-2
=>m=-1
đi qua = cắt.
<=> a khác a'
gọi đường thẳng: \(\left(3m-1\right)x+y=m-2\) là (d)
đk: 3m - 1 \(\ne0\)
=> \(m\ne\dfrac{1}{3}\)
có: \(\left\{{}\begin{matrix}3x+2y=5\\4x-3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5-2y}{3}\\y=\dfrac{4x-1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5-2y}{3}\\y=\dfrac{4.\dfrac{5-2y}{3}-1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5-2y}{3}\\y=\dfrac{17-8y}{3}.\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5-2y}{3}\\y=\dfrac{17-8y}{9}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5-2y}{3}\\17-8y-9y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5-2y}{3}\\17-17y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5-2y}{3}\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
thế x = 1 và y = 1 vào (d)
được: \(\left(d\right)=\left(3m-1\right).1+1-m+2=0\)
\(\Leftrightarrow3m-1+1-m+2=0\)
\(\Leftrightarrow2m+2=0\)
=> m = -1 (thỏa mãn điều kiện).
Vậy m = -1
