Cho các số hữu tỉ \(x=\dfrac{a}{b};y=\dfrac{c}{d}\left(a,b,c,d\in Z;b,d\ne0\right)\). Tổng \(x+y\) bằng
\(\dfrac{ac-bd}{bd}.\)\(\dfrac{ac+bd}{bd}.\)\(\dfrac{ad+bc}{bd}.\)\(\dfrac{ad-bc}{bd}.\)Hướng dẫn giải:Ta có \(x+y=\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad}{bd}+\dfrac{bc}{bd}=\dfrac{ad+bc}{bd}.\)
