Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)2a-3.-2+4=0=>2a+6+4=0=>2a=-10=>2=-5
b)3a-6-2.-1=2=>3a-6+2=2=>3a-6=0=>3a=6=>a=2
c)1-2.-3+-7-3a=-9=>1+6-7-3a=-9=>0-3a=-9=>3a=9=>a=3
b , 3a - b - 2c = 2 với b = 6 ; c = - 1
Ta có :
3a - 6 - 2 . ( - 1 ) = 2
3a - 8 . ( - 1 ) = 2
3a + 8 = 2
3a = 2 - 8
3a = - 6
=> a = - 6 : 3
a = - 2
\(a,\left(a-b+c\right)-\left(a+c\right)=a-b+c-a-c=-b\)
\(b,\left(a+b\right)-\left(b-a\right)+c=a+b-b+a+c=2a+c\)
\(c,-\left(a+b-c\right)+\left(a-b-c\right)=-a-b+c+a-b-c=-2b\)
\(d,a\left(b+c\right)-a\left(b+d\right)=ab+ac-ab-ad=ac-ad=a\left(c-d\right)\)
\(e,a\left(b-c\right)+a\left(d+c\right)=ab-ac+ad+ac=ab+ad=a\left(b+d\right)\)
a) (a - b + c) - (a + c)
= a - b + c - a - c
= (a - a) - b + (c - c)
= -b
b) (a + b) - (b - a) + c
= a + b - b + a + c
= 2a + (b - b) + c
= 2a + c
c) - (a + b - c) + (a - b - c)
= -a - b + c + a - b - c
= (-a + a) - (b + b) + (c - c)
= -2b
d) a(b + c) - a(b + d)
= ab + ac - ab - ad
= (ab - ab) + (ac - ad)
= ac - ad
= a(c - d)
e) a(b - c) + a(d + c)
= a(b - c + d + c)
= a[b - (c - c) + d]
= d(b + d)
Bài 1 :
\(a,\left(a-b\right)+\left(c-d\right)-\left(a-c\right)=-\left(b+d\right)\)
Ta có : \(VT=\left(a-b\right)+\left(c-d\right)-\left(a-c\right)\)
\(=a-b+c-d-a+c\)
\(=-\left(b+d\right)=VP\)
\(\Rightarrow\left(a-b\right)+\left(c-d\right)-\left(a-c\right)=-\left(b+d\right)\)
\(b,\left(a-b\right)-\left(c-d\right)+\left(b+c\right)=a+d\)
Ta có : \(VT=\left(a-b\right)-\left(c-d\right)+\left(b+c\right)\)
\(=a-b-c+d+b+c\)
\(=a+d=VP\)
\(\Rightarrow\left(a-b\right)-\left(c-d\right)+\left(b+c\right)=a+d\)
bài 1
ta có a+b = -4
=> a= -4-b
ta có b+c= -6
=> c= -6-b
ta có c+a =12
=> ( -4-b) + ( -6-b) = 12
=> -4-b - 6-b= 12
=> -10 - 2b = 12
=> -2b = -10 - 12
=> -2b = -22
=> b= 11
=> a= 4- 11= -15
=> c= 12- -15 = 17
1) (a-b+c)-(a+c)=-b=a-b+c-a-c=-b
2)(a+b)-(b-a)=2a+c=a+b-b+c+c=2c+a(ko the bang 2a+c)
3)-(a+b-c)+(a-b-c)=-a-b+c+c-b-c=-2c
tk minh nha
lam dc the thoi
\(A=\left(-2a+3b-4c\right)-\left(-2a-3b-4c\right)\)
\(a,=-2a+3b-4c+2a+3b+4c\)
\(=\left(-2a+2a\right)+\left(3b+3b\right)+\left(-4c+4c\right)\)
\(=0+\left(3b+3b\right)+0\)
\(=3b+3b=2.3b\)
\(b,\)Thay \(a=2012;b=-1;c=-2013\)vào biểu thức \(A\) ta có \(:\)
\(A=\left[-2.2012+3.\left(-1\right)-4.\left(-2013\right)\right]\)\(-\left[-2.\left(2012\right)-3.\left(-1\right)-4.\left(-2013\right)\right]\)
\(A=0\)
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)
a) Mình sửa lại 1 chút ở VP=-3b
Ta có: VT=-2(a+b-2c)+(2a-b-4c)
=-2a-2b+4c+2a-b-4c=-3b
=> VT=VP (đpcm)
b) Ta có VT=(a-b-c)-(a-b+c)=a-b-c-a+b-c=-2c
=> VT=VP (đpcm)
\(-\left(a-c\right)-\left(a-b+c\right)+\left(-d+c\right)\)
\(=-a-c-a+b-c-d+c\)
\(=-2a-c+b-d\)