Có \(a\left(b-c\right)-a\left(b+d\right)\)
\(\Rightarrow ab-ac-ab-ad\)
\(\Rightarrow ac+ad\)
\(\Rightarrow a\left(a+d\right)\)
Ta có: a.(b-c) - a.(b+d) = (ab - ac) - (ab+ad) = ab -ac - ab - ad
= ab - ab +ac+ad = ac+ad
Ta lại có: a(c+d) = ac+ad
Vậy a(b-c) - a(b+d) = a(c+d)
\(a\left(b-c\right)-a\left(b+d\right)\)
\(\Rightarrow ab-ac-ab-ad\)
\(\Rightarrow\left(ab-ab\right)+\left(ac+ad\right)\)
\(\Rightarrow a\left(c+d\right)\)
--> ĐPCM
