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 . a + b    = a ( a + b ) − b ( a + b ) = a 2 + ab − ba − b 2    = a 2 − b 2 = VP   ( dpcm )

VT = ( a + b ) ( a − b ) = a 2 − a . b + b . a − b 2 = a 2 − b 2 + ab − ab = a 2 − b 2 + 0 = a 2 − b 2 = VP . Vậy   ( a + b ) ( a − b ) = a 2 − b 2

VT = a + b 2 = a + b . a + b    = a ( a + b ) + b ( a + b ) = a 2 + ab + ba + b 2    = a 2 + 2 ab + b 2 = VP   ( dpcm )

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

VT = a ( b + c ) − b ( a − c ) = ab + ac − ba − bc    = ab + ac − ba + bc = ac + bc = c ( a + b ) = VP   ( dpcm )

a, \(a\left(b+c\right)-b\left(a-c\right)\)

\(=ab+ac-\left(ab-bc\right)\)

\(=ab+ac-ab+bc\)

\(=ac+bc\)

\(=\left(a+b\right)c\)

b,\(\left(a+b\right)\left(a-b\right)\)

\(=\left(aa+ab\right)-\left(ab+bb\right)\)

\(=aa+ab-ab-bb\)

\(=aa-bb\)

\(=a^2-b^2\)

a.(b+c)-b.(a-c)

=a.b+a.c-b.a+b.c

=(a.b-b.a)+a.b+b.c

=0+(a+b).c=(a+b).c(đpcm)

Ta có a .(b + c) - b .(a +c)

=a.b + a.c - b.a + b.c

=a.b - b.a + a.c + b.c

=0 + (a + b) . c

= (a +b) . c

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 $_$