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\(\text{ (a-b+c)-(a+c)}=a-b+c-a-c=\left(a-a\right)-b+\left(c-c\right)=-b\)
\(\left(a+b\right)-\left(b-a\right)+c=a+b-b+a+c=2a+c\)
\(-\left(a+b-c\right)+\left(a-b-c\right)=-a-b+c+a-b-c=-2b\)
\(a\left(b+c\right)-a\left(b+d\right)=ab+ac-ab+ad=ac+ad=a\left(c+d\right)\)
\(a\left(b-c\right)+a\left(d+c\right)=a\left(b-c+d+c\right)=a\left(b+d\right)\)
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 = -b
2. (a+b) - (b-a) +c = a+b -b +a +c =2a+c
3. -(a+b-c)+(a-b-c) = -a-b+c a-b-c = -2b
4. a(b+c) -a(b+d) = a(b+c-b-d) = a( c-d)
5. a(b-c) +a(d+c) = a(b-c+d+c) = a(b+d)
1.= a-b+c-a-c= (a-a)-b+(c-c)=0-b+0=-b
2.=a+b-b+a+c=a+a+b-b+c=2a+c
3.=-a-b+c+a-b-c=-a+a-(b+b)+c-c=-2b
4.=ab+ac-ab-ad=ac-ad=a(c-d)
5.=ab-ac+ad+ac=(-ac+ac)+ab+ad=ab+ad=a(b+d)
tk mik nha, chúc bn học tốt
1/ (a-b+c) - (a+c) = a-b+c-a-c = -b
2/ (a+b) - (b-a) + c = a + b - b + a + c = 2a + c
3/ - (a+b-c) + (a-b-c) = -a - b + c + a - b - c = - 2b
4/ a(b+c) - a(b+d) = a(b+c-b-d) = a(c-d)
5/ a(b-c) + a(d+c) = a(b-c+d+c) = a(b+d)
1/ (a-b+c) - (a+c) = a-b+c -a-c
= (a-a) +(c-c) -b
= 0+0-b
= -b
2/ (a+b) - (b-a) +c= a+b -b+a +c
= (a+a)+ (b-b) +c
= 2a+c
3/ -(a+b-c) + (a-b-c) = -a-b+c + a-b-c
= (-a+a) + (-b-b) +(c-c)
= 0 +(-2b)+ 0
= -2b
4/ a (b+c)- a(b+d) = ab+ac - ab-ad
= (ab-ab) +(ac-ad)
= 0+ a.(c-d)
= a.(c-d)
5/ a(b-c) + a(d+c) = ab-ac + ad +ac
= (-ac+ac) + (ab+ad)
= 0+ a.(b+d)
= a.(b+d)
mấy cái này chỉ cần phá dấu ngoặc rồi rút gọn là ra thui
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
=(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)
Bai 1 :
a, Ta có : \(\left(a-b+c\right)-\left(a+c\right)=a-b+c-a-c=-b\) ( đpcm )
b, Ta có : \(\left(a+b\right)-\left(b-a\right)+c=a+b-b+a+c=2a+c\) (đpcm)
c, Ta có : \(-\left(a+b-c\right)+\left(a-b-c\right)=-a-b+c+a-b-c\)
= \(-2b\) ( đpcm )
d, Ta có : \(a\left(b+c\right)-a\left(b+d\right)=ab+ac-ab-ad\)
= \(ac-ad=a\left(c-d\right)\) ( đpcm )
e, Ta có : \(a\left(b-c\right)+a\left(d+c\right)=ab-ac+ad+ac\)
= \(ab+ad=a\left(b+d\right)\) ( đpcm )
đpcm là gì vậy??