Question

chứng tỏ rằng

1, a(b+c)-b(a-c)=(a+b)c

2, a(b-c)-a(b+d)=-a(c+d)

3, (a+b)(c+d)-(a+d)(b+c)=(a-c)(d-b)

Answer 1

1, a(b+c)-b(a-c)=(a+b)c

\(ab+ac-ba+bc=\left(a+b\right)c\)

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

\(a.0+\left(a+b\right)c=\left(a+b\right)c\)

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

\(\Rightarrowđpcm\)

2, a(b-c)-a(b+d)=-a(c+d)

\(ab-ac-ab-ad=a.\left(c+d\right)\)

\(a.\left(b-c-b-d\right)=a\left(-c-d\right)\)

\(a.\left(-c-d\right)=a.\left(-c-d\right)\)

\(\Rightarrowđpcm\)

3, (a+b)(c+d)-(a+d)(b+c)=(a-c)(d-b)

=ac+ad+bc+bd-ab-ac-bd-dc

=ad-ab+bc-dc

=(ad-ab)+(bc-dc)

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

=a(d-b)-c(d-b)

=(a-c)(d-b) =VP.

\(\Rightarrowđpcm\)

học tốt

Answer 2

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

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

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

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

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

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

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

=a.b-a.c-a.b-a.d

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

=0-a.c-a.d

=-a.c-a.d

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

=-a.(c+d)

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

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

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

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

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

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