a(b-c)-a(b+d)=-a(c+d)
<=> ab-ac-ab-ad=-ac-ad
<=> (ab-ab)+-ac-ad=-ac-ad
<=> 0-ac-ad=-ac-ad
<=>-ac-ad=-ac-ad (đpcm)
a(b-c)-a(b+d)=-a(c+d)
Ta có : a(b-c)-a(b+d)
= ab - ac - ab - ad
= -ac - ad
= -a( c + d ) \(\rightarrow\)ĐPCM
# HOK TỐT #
Ta có :
a(b-c) - a(b+d) = a(b-c-b-d) = a(-c-d)=-a(c+d)
a(b-c)-a(b+d)=-a(c+d)
<=>a(b-c)-a(b+d)+a(c+d)=0
<=>a(b-c-b-d+c+d)=0......
còn lại bn tự lm nha
học tốt
Trả lời:
\(a\left(b-c\right)-a\left(b+d\right)=-a\left(c+d\right)\)
\(\Leftrightarrow ab-ac-ab-ad=-ac-ad\)
\(\Leftrightarrow-ac-ad=-ac-ad\)\(\left(đpcm\right)\)
P/s: Mk không hiểu đề bài cho lắm...
Hok tốt!
Good girl
\(a\left(b-c\right)-a\left(b+d\right)=-a\left(c+d\right)\)
\(ab-ac-ab-ad=-ac-ad\)
\(-ac-ad=-ac-ad\left(đpcm\right)\)
