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\(\text{ (a-b+c)-(a+c)}=a-b+c-a-c=\left(a-a\right)-b+\left(c-c\right)=-b\)
\(\left(a+b\right)-\left(b-a\right)+c=a+b-b+a+c=2a+c\)
\(-\left(a+b-c\right)+\left(a-b-c\right)=-a-b+c+a-b-c=-2b\)
\(a\left(b+c\right)-a\left(b+d\right)=ab+ac-ab+ad=ac+ad=a\left(c+d\right)\)
\(a\left(b-c\right)+a\left(d+c\right)=a\left(b-c+d+c\right)=a\left(b+d\right)\)
C1
B = (a + b - c)-(b + c - a)-(a - b - c)
=a+b-c-b-c+a-a+b+c
=a+b-c
C2
ta có:
a) (a-b)+(c-d)
=a-b+c-d
=a+c-b-d
=(a+c)-(b+d)
vậy .....
b)ta có:
(a-b)-(c-d)
=a-b-c+d
=a+d-b-c
=a+d-b-c
=(a+d)-(b+c)
1,(a-b+c)-(a+c)=a-b+c-a-c=-b
2,(a+b)-(b-c)+c=a+b-b+c+c=2c+a (không thể bằng 2a+c đc)
3,-(a+b-c)+(a-b-c)=-a-b+c+a-b-c=-2b
1) (a-b+c)-(a+c)=a-b+c-a-c=-b (đpcm)
2) (a+b)-(b-a)+c=a+b-b+a+c=2a+c (đpcm)
3) -(a+b-c)+(a-b-c)=-a-b+c+a-b-c=-2b (đpcm)
4) a(b+c) -a(b+d)=ab+ac-ab-ad=ac-ad=a(c-d) (đpcm)
5) a(b-c)+a(d+c)=ab-ac+ad+ac=ab+ad=a(b+d) (đpcm)
CHÚC BẠN HỌC TỐT NHÉ!
\(\left(a-b+c\right)-\left(a+c\right)=-b\)
\(a-b+c-a-c=-b\)
\(-b=-b\left(đpcm\right)\)
\(\left(a+b\right)-\left(b-a\right)+c=2a+c\)
\(a+b-b+a+c=2a+c\)
\(2a+c=2a+c\left(đpcm\right)\)
\(-\left(a+b-c\right)+\left(a-b-c\right)=-2b\)
\(-a-b+c+a-b-c=-2b\)
\(-2b=-2b\left(đpcm\right)\)
lm cx dễ thoi , bn lm tiếp nha !
1,(a-b+c)-(a+c)
=a-b+c-a-c
=(a-b)-b+(c-c)
=0-b+0
=-b(đpcm)
2,(a+b)-(b-a)+c
=a+b-b+a+c
=(a+a)+(b-b)+c
=2a+0+c
=2a+c(đpcm)
3,-(a+b-c)+(a-b-c)
=-a-b+c+a-b-c
=(-a+a)-b-b+(c-c)
=0-b-b+0
=-b-b
=-2b(đpcm)
4,a(b+c)-a(b+d)
=ab+ac-ab+ad
=(ab-ab)+ac+ad
=0+ac+ad
=ac+ad
=a(c+d)(đpcm)
5,a(b-c)+a(d+c)
=ab-ac+ad+ac
=(-ac+ac)+ab+ad
=0+ab+ad
=ab+ad
=a(b+d)(đpcm)
1. ( a - b + c ) - ( a + c )
=> a - b + c - a - c
=> ( a - a ) - b + ( c - c )
=> 0 - b + 0
=> -b ( đpcm ).
2. ( a + b ) - ( b - a ) + c
=> a + b - b + a + c
=> ( a + a ) + ( b - b ) + c
=> 2a + c ( đpcm )
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
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)