ta có:a(b−c)−a(b+d)=−a(c+d)
VT(vế trái)=a(b−c)−a(b+d)
=ab−ac−ab−ad
=(ab−ab)−ac−ad
=0−a(c+d)
=−a(c+d)=VP(vế phải)
\(a\left(b-c\right)-a\left(b+d\right)\)
\(=a\left(b-c-b-d\right)\)
\(=a\left(-c-d\right)\)
\(=-a\left(c+d\right)\left(dpcm\right)\)
Ta có: a(b-c)-a(b+d)
=ab-ac-ab-ad
=-ac-ad=-(ac+ad)=-a(c+d)
Vì -a(c+d)=-a(c+d) nên a(b-c)-a(b+d)=-a(c+d)
ta có:a(b−c)−a(b+d)=−a(c+d)
⇒VT(vế trái)=a(b−c)−a(b+d)
=ab−ac−ab−ad
=(ab−ab)−ac−ad
=0−a(c+d)
=−a(c+d)=VP(vế phải)
vậy a(b-c)-a(b+d)=-a(c+d)
Ta có: \(a\left(b-c\right)-a\left(b+d\right)\)
\(=a\left(b-c-b-d\right)\)
=a(-c-d)
=-a(c+d)(đpcm)
