Giải thích các bước giải:
$1,$
$d,a(b + c) - a(b + d)$
$= ab + ac - ab - ad $
$= ac - ad $
$= - a(d - c) $
$Vậy$ $ a(b+c) - a(b + d) \neq -a(c+d)$
Sửa đề :
$a(b - c) - a(b + d)$
$= ab - ac - ab - ad$
$= - ac - ad $
$= - a(c + d)$
$Vậy$ $a(b-c) - a(b + d) = - a(c+d)$
$e, (a+b)(c + d) - (a + d)(b - c)$
$= ac + ad + bc + bd - (ab - ac + bd - cd)$
$= ac + ad + bc + bd - ab + ac - bd + cd$
$= 2ac - ab + ad + bc + cd$ $(đề sai)$
Sửa đề :
$(a+b)(c+d) - (a + d)(b + c)$
$= ac + ad + bc + bd - (ab + ac + bd + cd)$
$= ac + ad + bc + bd - ab - ac - bd - cd$
$= ad + bc - ab - cd$
$= (ad - cd) + (bc - ab)$
$= d(a-c) + b(c - a)$
$= d(a - c) - b(a - c)$
$= (a - c)(d - b)$
$Vậy$ $(a+b)(c+d) - (a + d)(b + c) = (a-c)(d-b)$
