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VT=(a+b)(a-b)=a(a-b)+b(a-b)=a2-ab+ab-b2=a2-b2
ta có: VT=VP=>đpcm
(a+b)(a-b)=(a+b).a-(a+b).b
=(a2+ab)-(ab+b2)
=a2+ab-ab-b2
=a2-b2
a/ VT = \(a\left(b+c\right)-b\left(a-c\right)=ab+ac-ba+bc=ac+bc\)
\(=c\left(a+b\right)\) = VP => ĐPCM
b/ VT = \(\left(a+b\right)\left(a-b\right)=a^2-ab+ab-b^2=a^2-b^2\)= VP
=> ĐPCM
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
\(a\left(b-c\right)-a\left(b+d\right)=-a\left(c+d\right)\)
\(\Leftrightarrow a[\left(b-c\right)-\left(b+d\right)]=-a\left(c+d\right)\)
\(\Leftrightarrow a\left(b-c-b-d\right)=-a\left(c+d\right)\)
\(\Leftrightarrow a\left(-c-d\right)=-a\left(c+d\right)\)
\(\Leftrightarrow-a\left(c+d\right)=-a\left(c+d\right)\)
vậy ...
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)+c(a-b)=b(a-c)
ab - ac + ca - cb - ba + cb = 0
(ab - ab) - (bc - bc) - (ac - ac) =0
0 - 0 -0 = 0 (đpcm)
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,\(\left(a+b\right)\left(a-b\right)\)
\(=\left(aa+ab\right)-\left(ab+bb\right)\)
\(=aa+ab-ab-bb\)
\(=aa-bb\)
\(=a^2-b^2\)