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a, Có : a.(b-c)+c.(a-b) = ab-ac+ac-bc = ab-bc = b.(a-c)
b, Có : (a+b).(c+d)-(a+d).(b+c) = ac+bc+ad+bd-ab-bd-ac-cd = ad+bc-ab-cd
= (ad-cd)-(ab-bc) = d.(a-c)-b.(a-c) = (a-c).(d-b)
Tk mk nha
a, Có : a.(b-c)+c.(a-b) = ab-ac+ac-bc = ab-bc = b.(a-c)
b, Có : (a+b).(c+d)-(a+d).(b+c) = ac+bc+ad+bd-ab-bd-ac-cd = ad+bc-ab-cd
= (ad-cd)-(ab-bc) = d.(a-c)-b.(a-c) = (a-c).(d-b)
tk cho mk nha $_$
KHAI TRIEN VE TRAI
(a+b)(c+d)-(a+d)(b+c)
=ac+ad+bc+bd-ab-ac-bd-cd
=ad+bc-ab-cd
=a(-b+d)-c(-b+d)
=(a-c)(-b+d) GIONG NHU VE PHAI
THAY DUNG THI
\(a\left(b-c\right)-a\left(b+d\right)=-a\left(c+d\right)\)
\(\Leftrightarrow a[\left(b-c\right)-\left(b+d\right)]=-a\left(c+d\right)\)
\(\Leftrightarrow a\left(b-c-b-d\right)=-a\left(c+d\right)\)
\(\Leftrightarrow a\left(-c-d\right)=-a\left(c+d\right)\)
\(\Leftrightarrow-a\left(c+d\right)=-a\left(c+d\right)\)
vậy ...
a) Ta có: \(a\left(b-c\right)+c\left(a-b\right)\)
\(=ab-ac+ac-bc\)
\(=ab-bc\)
\(=b\left(a-c\right)\)(đpcm)
b) Ta có: \(a\left(b-c\right)-b\left(a+c\right)\)
\(=ab-ac-ab-bc\)
\(=-ac-bc\)
\(=\left(-c\right)\cdot\left(a+b\right)=\left(a+b\right)\cdot\left(-c\right)\)(đpcm)
c) Ta có: \(a\left(b+c\right)-b\left(a-c\right)\)
\(=ab+ac-ab+bc\)
\(=ac+bc\)
\(=\left(a+b\right)\cdot c\)(đpcm)
d) Sửa đề: \(a\left(b-c\right)-a\left(b+d\right)\)
Ta có: \(a\left(b-c\right)-a\left(b+d\right)\)
\(=ab-ac-ab-ad\)
\(=-ac-ad\)
\(=-a\left(c+d\right)\)(đpcm)
VT = a ( b + c ) − b ( a − c ) = ab + ac − ba − bc = ab + ac − ba + bc = ac + bc = c ( a + b ) = VP ( dpcm )
a, a(b+c)−b(a−c)a(b+c)−b(a−c)
=ab+ac−(ab−bc)=ab+ac−(ab−bc)
=ab+ac−ab+bc=ab+ac−ab+bc
=ac+bc=ac+bc
=(a+b)c=(a+b)c
b,(a+b)(a−b)(a+b)(a−b)
=(aa+ab)−(ab+bb)=(aa+ab)−(ab+bb)
=aa+ab−ab−bb
VT = a ( b + c ) − b ( a − c ) = ab + ac − ba + bc = ( ab − ab ) + ( ac + bc ) = 0 + a + b . c = VP Vậy a ( b + c ) − b ( a − c ) = ( a + b ) . c
a.(b+c)-b.(a-c)=(a+b).c
=ab+ac-ba+bc
=ac+bc
=(a+b).c