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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) \(A=2b=2.\left(-1\right)=-2\)
![](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)
(a + b + c) . (a + b + c) - 2.(a . b + b . c + c . a)
= a . a + a . b + a . c + b . a + b . b + b . c + c . a + c . b + c . c - 2ab - 2bc - 2ca
= a2 + ab + ac + ab + b2 + bc + ac + bc + c2 - 2ab - 2bc - 2ca
= a2 + b2 + (ab + ab) + (bc + bc) + (ac + ac) - 2ab - 2bc - 2ca
= a2 + b2 + 2ab + 2bc + 2ca - 2ab - 2bc - 2ca
= a2 + b2.
![](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 + b + c) . ( a + b + c ) - 2 . ( a . b + b.c + c.a )
= a
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