\(a,VT=ab-ac+ac-bc=ab-bc=b\left(a-c\right)=VP\\ b,VT=ab-ac-ab-bc=-ac-bc=\left(a+b\right)\left(-c\right)=VP\)

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

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

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

\(=aa-bb\)

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

bai toan nay khó

VT=(a+b)(a-b)=a(a-b)+b(a-b)=a²-ab+ab-b²=a²-b²

ta có: VT=VP=>đpcm

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 + 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(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

tl thi ko tl con doi tich

toàn đòi tic ko chẳng ra làm sao

a) a(b+c)-b(a-c)=a.b+a.c-b.a-b.c=a.b-b.a+a.c-b.c=0+a.c-b.c=(a+b)c

=> a(b+c)-b(a-c)=(a+b)c