Ta có: \(\left\{{}\begin{matrix}x_1=45t\\x_2=240-36t\end{matrix}\right.\)
Để 2 xe gặp nhau thì \(x_1=x_2\)
\(\Leftrightarrow45t=240-36t\)
\(\Leftrightarrow t=\dfrac{80}{27}\)
Vậy hai xe gặp nhau lúc: \(6h30p+\dfrac{80}{27}h\approx9h30p\)
Cách A: \(45\cdot\dfrac{80}{27}\approx133,\left(3\right)km\)