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A=(a-b)+(a+b+c)-(a-b-c)
=a-b+a+b+c-a+b+c
=(a+a-a)+(-b+b+b)+(c+c)
= a+b+c.2
= a+b+2c
B=(a-b)-(b-c)+(c-a)-(a-b-c)
=a-b-b+c+c-a-a+b+c
=(a-a-a)+(-b-b+b)+(c+c+c)
= (-a)+ (-b) +c.3
= (-a)+(-b)+3c
C=(-a+b+c)-(a-b+c)-(a+b-c)
= (-a)+b+c-a+b-c-a-b+c
=(-a-a-a)+(b+b-b)+(c-c+c)
= (-a.3) +b+c
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)
a, A = (a - b) + (a + b - c) - (a - b - c)
= a - b + a + b - c - a + b + c
= a + b
b, B = (a - b) - (b - c) + (c - a) - (a - b - c)
= a - b - b + c + c - a - a + b + c
= - a - b + 3c
c, C = ( - a + b + c) - (a - b + c) - ( -a + b - c)
= -a + b + c - a + b - c + a - b + c
= -a + b + c
\(a,A=\left(a-b\right)+\left(a+b-c\right)-\left(a-b-c\right)\)
\(=a-b+a+b-c-a+b+c\)
\(=a+b\)
\(b,B=\left(a-b\right)-\left(b-c\right)+\left(c-a\right)-\left(a-b-c\right)\)
\(=a-b-b+c+c-a-a+b+c\)
\(=-a-b+3c\)
\(c,C=\left(-a+b+c\right)-\left(a-b+c\right)-\left(-a+b-c\right)\)
\(=-a+b+c-a+b-c+a-b+c\)
\(=-a+b+c\)
a. A = (a-b)+(a+b-c)-(a-b-c)=a-b+a+b-c-a+b+c=(a-a+a) + (b-b+b)+(c-c)=4+3+0=4+b
b.B=(a-b)-(b-c)+(c-a)-(a-b-c)=a-b-b+c+c-a-a+b+c=(a-a-a)+(b-b-b)+(c+c+c)=-a+(-b)+3c
c.C=(-a+b+c)-(a-b+c)-(-a+b-c)=-a+b+c+b-a+b-c-a-b+c= -(a+a+a)+(b+b+b)+(c+c-c)=-3a+3b+c
****
a) A=(a-b)+(a+b-c)-(a-b-c)=a+b
b) B=(a-b)-(b-c)+(c-a)-(a-b-c)=-a-b+3c
c) C=(-a+b+c)-(a-b+c)-(-a+b-c)=-a+b+c
A = ( a - b ) + ( a + b - c ) - ( a - b - c )
A = a - b + a + b - c - a + b + c
A = a + b
B = ( a - b ) - ( b - c ) + ( c - a ) - ( a - b - c )
B = a - b - b + c + c - a - a + b + c
B = -a - b + 3c
C = ( -a + b + c ) - ( a - b + c ) - ( -a + b - c )
C = -a + b + c - a + b - c + a - b + c
C = -a + b + c
\(D=\left(a+b-c\right)-\left(a-b+c\right)+\left(b+c-a\right)-\left(a-b-c\right)\)
\(D=a+b-c-a+b-c+b+c-a-a+b+c\)
\(D=\left(a-a-a-a\right)+\left(b+b+b+b\right)+\left(c+c-c-c\right)\)
\(D=4b-3a\)
a, \(A=\left(a-b\right)+\left(a+b-c\right)-\left(a-b-c\right)\)
\(=a-b+a+b-c-a+b+c=a+b\)
b, \(B=\left(a-b\right)-\left(b-c\right)+\left(c-a\right)-\left(a-b-c\right)\)
\(=a-b-b+c+c-a-a+b+c=-a-b+3c\)
c, \(C=\left(-a+b+c\right)-\left(a-b+c\right)-\left(-a+b-c\right)\)
\(=-a+b+c-a+b-c+a-b+c=-a+b+c\)
(a+b+c)-(a-b+c)= a+b+c-a+b-c=(a-a)+(b+b)+(c-c)=2b
(a+b-c)+(a-b)-(a-b-c)=a+b-c+a-b-a+b+c= (a+a-a)+(b-b+b)+(c-c)=a+b
-(a-b-c)+(-a+b-c)-(-a-b+c)=-a+b+c-a+b-c+a+b-c= (-a-a+a)+(b+b+b)+(c-c-c)=a+3b-c
(a + b + c) - (a - b + c)
= a+ b + c - a + b - c
= (a - a) + (b + b) + (c - c) = 2b
Tương tự
Ta có:
A=a-b+a+b-c-a+b+c
=a+b
B=a-b-b+c+c-a-a+b+c=3c-a-b
C=(-a+b+c)-(a-b+c)-(-a+b-c)=-a+b+c-a+b-c+a-b+c=-a+b+c