\(ab-cd-\left(ad+bc\right)=\left(ab-bc\right)-\left(cd+ad\right)=b\left(a-c\right)-d\left(a+c\right)\)
quên
\(ab+cd-ad-bc=\left(ab-bc\right)+\left(cd-ad\right)=b\left(a-c\right)+d\left(c-a\right)=\left(b-d\right)\left(a-c\right)⋮a-c\Rightarrow ad+bc⋮a-c\)
Ta có : \(ab+cd-\left(ad+bc\right)=ab-ad+cd-bc\)
\(=a.\left(b-d\right)-c.\left(b-d\right)\)
\(=\left(a-c\right).\left(b-d\right)⋮a-c\) (1)
Có : \(ab+cd⋮a-c\) (2)
Từ (1) và (2) \(\Rightarrow ad+bc⋮a-c\)
* Áp dụng công thức : \(\)\(\left\{{}\begin{matrix}a-b⋮m\\a⋮m\end{matrix}\right.\) \(\Rightarrow b⋮m\)