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A = -{-(a + b) - [-(a - b) - (a + b)]}
= -{-a - b - [-a + b - a - b]}
= -[-a - b - (-2a)]
= -(-a - b + 2a)
= -(a - b)
= -a + b
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\left(-a+b-c\right)+\left(a-b\right)-\left(a-b+c\right)\)
\(=-a+b-c+a-b-a+b-c\)
=-a+b-2c
![](https://rs.olm.vn/images/avt/0.png?1311)
B1:
a, a+b+(-a)+b+a+(-c)+(-a)+(-c)=[a+(-a)+a+(-a)]+(b+b)+[(-c)+(-c)]=0+2.b+(-2).c
b, a+b+(-c)+a+(-b)+c+(-b)+(-c)+a+(-a)+b+c=[a+a+a+(-a)]+[b+(-b)+(-b)+b]+[(-c)+c+(-c)+c]=2.a+0+0=2a
B2:
N=(a+b)-(a-b)+(a+b)=a+b+(-a)+b+a+b=[a+(-a)+a)+(b+b+b)=a+3.b
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rút gọn biểu thức
a,A=(a-b)-(a-b+c)
b,B=-(a+b+c)-(a+b-5)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =b-c-a+c+1-a-b+c
=-2a+1
b: =a-b-c-b+c+a+c-b-a
=c-3b+a
c: =2(a-b-b+c-c+a)
=2(2a-2b)
=4a-4b
a) \(\left(b-c\right)-\left(a-c-1\right)-\left(a+b-c\right)\)
\(=b-c-a+c+1-a-b+c\)
\(=c-2a+1\)
b) \(\left(a-b-c\right)-\left(b-c-a\right)+\left(c-b-a\right)\)
\(=a-b-c-b+c+a+c-b-a\)
\(=a-3b+c\)
c) \(2\cdot\left(a-b\right)-2\cdot\left(b-c\right)-2\cdot\left(c-a\right)\)
\(=2\cdot\left(a-b-b+c-c+a\right)\)
\(=2\cdot\left(2a-2b\right)\)
\(=4a-4b\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 :
\(A=\left(-a+b-c\right)-\left(-a-b-c\right)\)
\(=-a+b-c+a+b+c=2b\)
Ta có b = -1 ta được : \(2b=2\left(-1\right)=-2\)
Vậy \(A=-2\)
\(B=\left(-2a+3b-4c\right)-\left(-2a-3b-4c\right)=-2a+3b-4c+2a+3b+4c\)
\(=6b\)
Ta có : b = -1 khi đó: \(B=6b=6\left(-1\right)=-6\)
Vậy B = -6
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có:
\(A=\left(-a+b-c\right)-\left(-a-b-c\right)\)
\(=-a+b-c+a+b+c\)
\(=\left(-a+a\right)+\left(b+b\right)+\left(-c+c\right)\)
\(=0+2b+0\)
\(=2b\)
b) \(A=2b=2.\left(-1\right)=-2\)
a)
\(=a^2+ab+ab+b^2\)
\(=a^2+2ab+b^2\)
\(=\left(a+b\right)^2\)
b)
\(=a^2-ab-ab+b^2\)
\(=a^2-2ab+b^2\)
\(=\left(a-b\right)^2\)
a) 2(a+b)
b) 2(a-b)