Chứng tỏ
a) (a-b+c)-(a+c)=-b
b) (a+b)-(b-a)+c=2a+c
c) -(a+b-c) +(a-b-c) =-2b
d) a(b+c)-a(b+d)=a(c-d)
e) a(b-c)+a(d+c)=a(b+d)
1.Chứng tỏ
a) ( a - b + c ) - ( a +c ) = - b
b) ( a + b) - ( b - a ) + c = 2a + c
c) -(a + b - c) + (a - b - c) = -2b
d) a(b + c) - a( b + d) = a(c - d)
e) a(b - c) + a(d + c)= a(b +d)
chứng tỏ:
a)(a-b+c)-(a+c)=-b
b) (a+b) -(b-a) + c =2a +c
c) -(a+b-c)+(a-b-c)=-2b
d) a(b+c)-a(b+d) =a(c-d)
e) a(b-c) +a(d+c) =a(b+d)
a)(a-b+c)-(a+c)
=a-b+c-a-c=a-a+c-c-b
=0-b=-b
b)(a+b)-(b-a)+c
=a-b-b+a+c=2a+c
Chứng tỏ:
a, ( a - b + c ) - ( a + c ) = -b
b, ( a + b ) - ( b - a ) + c = 2a + c
c, - ( a + b - c ) + ( a - b - c ) = -2b
d, a( b + c ) - a( b + d ) = a( c - d )
e, a( b - c ) + a( d + c ) = a( b + d )
chứng tỏ
a,(a-b+c)-(a+c)=-b
b,(a+b)-(b-a)+c=2a+c
c,-(a+b-c)+(a-b-c)=-2b
d.a(b+c)-a(b+d)=a(c-d)
e,a(b-c)+a(d+c=a(b+d)
bài này bn áp dụng quy tắc bỏ ngoặc & tính chất của tổng đại số là đc!
a,(a-b+c)-(a+c)
= a - b + c - a - c
= (a-a) + (c-c) - b
= 0 + 0 - b
= -b
=> (a-b+c)-(a+c) = - b
b,(a+b)-(b-a)+c
= a + b - b + a + c
= (a+a) + (b - b) + c
= 2a + 0 + c
= 2a + c
=> (a+b)-(b-a)+c = 2a + c
c,-(a+b-c)+(a-b-c)
= -a - b + c + a - b - c
= (a-a) - (b+b) + (c+c)
= 0 - 2b + 0
= -2b
=> -(a+b-c)+(a-b-c) = -2b
d, a(b+c)-a(b+d)
= ab + ac - ab + ad
= (ab-ab) + (ac - ad)
= 0 + a(c-d)
= a(c-d)
=> a(b+c)-a(b+d) = a(c-d)
e,a(b-c)+a(d+c=a(b+d) => câu này mk ko hiểu đề
a) (a - b + c) - (a + c)
= a - b + c - (a + c)
= (a + c) - (a + c) - b
= 0 - 0 - b
= -b
a, (a-b+c)-(a+c)=a-b+c-a-c=-b
b, (a+b)-(b-a)+c=a+b-b+a+c=2a+c
c, -(a+b-c)+(a-b-c)=-a-b+c+a-b-c=-2b
d, a(b+c)-a(b+d)=a(b+c-b-d)=a(c-d)
e, a(b-c)+a(d+c)=a(b-c+d+c)=a(b+d)
Chứng tỏ:
a) (a-b+c)-(a+c)=-b
b) (a+b)-(b-a)+c=2a+c
c)-(a+b-c)+(a-b-c)=-2b
d) a(b+c)-a(b+d)=a(c-d)
e)a(b-c)+a(d+c)=a(b+d)
a, (a-b+c)-(a+c)=-b
<=>a-b+c-a-c=-b
<=>(a-a)+(c-c)-b=-b
<=>0+0-b=-b
<=>-b=-b
Vậy (a-b+c)-(a+c)=-b
b) (a+b)-(b-a)+c=2a+c
<=>a+(b-b)+a+c=2a+c
<=>a+a+c=2a+c
<=>2a+c=2a+c
Vậy (a+b)-(b-a)+c=2a+c
c) -(a+b-c)+(a-b-c)=-2b
<=>-a-b+c+a-b-c=-2b
<=>(-a+a)+(c-c)-(b+b)=-2b
<=>0+0-2b=-2b
<=>-2b=-2b
Vậy -(a+b-c)+(a-b-c)=-2b
d) a(b+c)-a(b+d)=a(c-d)
<=>ab+ac-ab-ad=a(c-d)
<=>a(b+c-b-d)=a(c-d)
<=>a(c-d)=a(c-d)
Vậy a(b+c)-a(b+d)=a(c-d)
e) a(b-c)+a(c+d)=a(b+d)
<=>ab-ac+ac+ad=a(b+d)
<=>a(b-c+c+d)=a(b+d)
<=>a(b+d)=a(b+d)
Vậy a(b-c)+a(c+d)=a(b+d)
thang le ngoc anh bi ngu ak
Giúp mk nhé:
Chứng tỏ:
a) (a-b+c)-(a+c)= -b
b) (a+b)-(b-a)+c= 2a+c
c) -(a+b-c)+(a-b-c)= -2b
d) a(b+c)-a(b+d)= a(c-d)
e) a(b-c)+a(d+c)= a(b+d)
a) (a-b+c)-(a+c)
= a-b+c-a-c
=-b
b) (a+b)-(b-a)+c
=a+b-b+a+c
=2a+c
c) -(a+b-c)+(a-b-c)
=-a-b+c+a-b-c
=-2b
d) a(b+c)-a(b+d)
=ab+ac-ab-ad
=ac-ad
=a(c-d)
e) a(b-c)+a(d+c)
= ab-ac+ad+ac
=ab+ad
=a(b+d)
\(\text{ (a-b+c)-(a+c)}=a-b+c-a-c=\left(a-a\right)-b+\left(c-c\right)=-b\)
\(\left(a+b\right)-\left(b-a\right)+c=a+b-b+a+c=2a+c\)
\(-\left(a+b-c\right)+\left(a-b-c\right)=-a-b+c+a-b-c=-2b\)
\(a\left(b+c\right)-a\left(b+d\right)=ab+ac-ab+ad=ac+ad=a\left(c+d\right)\)
\(a\left(b-c\right)+a\left(d+c\right)=a\left(b-c+d+c\right)=a\left(b+d\right)\)
1. Chứng tỏ :
a). (a - b + c) - (a + c) = -b
b). (a + b) - (b - a) + c = 2a + c
c). - (a + b - c) + (a - b - c) = -2b
d). a(b+ c) - a(b + d) = a(c - d)
e). a(b - c) + a(d + c) = a(b + d)
Mong các bạn giúp mình ! Mình đang cần gấp lắm! Mình xin cảm ơn !
Chứng tỏ:
(a-b+c)-(a+c)=-b(a+b)-(b-a)+c=2a+c-(a+b-c)+(a-b-c)=-2ba(b+c-a(b+d)=a(b+d)a(b-c)+a(d+c)=a(b+d)1 (a-b+c)-(a+c) = a-b+c-a-c
=(a-a)+(c-c)-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)
=-2b
4 MÌNH KHÔNG HIỂU ĐỀ, BẠN GHI ĐỀ RÕ LẠI 1 CHÚT
5 a(b-c)+ a(d+c) = a(b-c+d+c)
= a(b-c+d+c)
= a( b+d)
Giải thích các bước giải:
1:a-b+c)-(a+c)=a-b+c-a-c=-b
2:(a+b)-(b-a)+c=a+b -b +a=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+a.c - a.b-a.d=a(c-d)
5/a.(b-c)+a.(d+c)= a.b-a.c+a.d+a.c=a(b+d)