ta có: -(-a+b+c)+(b+c-1)= a-b-c+b+c-1=a-1   (1)

         (b-c+6)-(7-a+b)+c= b-c+6-7+a-b+c=a-1  (2)

Từ (1),(2) => -(-a+b+c)+(b+c-1)=(b-c+6)-(7-a+b)+c

Vế trái = -(-a+b+c)+(b+c-1)

= a-b-c+b+c-1

= a+(-b+b)+(-c+c)-1

= a+0+0-1

= a-1

Vế phải = (b-c+6)-(7-a+b)+c

= b-c+6-7+a-b+c

= (b-b)+(-c+c)+(6-7)+a

= 0+0-1+a

= a-1

- Vậy -(-a+b+c)+(b+c-1)=(b-c+6)-(7-a+b)+c

a)

Có: -(-a + b + c) + (b + c - 1) = a - b - c + b + c - 1

                                                = a - 1

Lại có: (b - c + 6) - (7 - a + b) + c = b - c + 6 - 7 + a - b + c 

                                                      = a - 1

Vì a - 1 = a - 1

nên -(-a + b + c) + (b + c - 1) = (b - c + 6) - (7 - a + b) + c (đpcm)

VT=\(-\left(-a+b+c\right)+\left(b+c-1\right)\)

\(=a-b-c+b+c-1\)

=a-1

\(VP=\left(b-c+6\right)-\left(7-a+b\right)+c\)

\(=b-c+6-7+a-b+c\)

=a-1

=>VT=VP

=>\(-\left(-a+b+c\right)+\left(b+c-1\right)=\left(b-c+6\right)-\left(7-a+b\right)+c\)

a) -(-a+b+c)+(b+c-1)=(b-c+6)-(7-a+b)+c

VP=(b-c+6)-(7-a+b)+c

=b-c+6-7+a-b+c

=b-c-1+a-b+c

=(b+c-1)+(a-b-c)

=(b+c-1)-(-a+b+c)=VT(đccm)

b) A+B=​a+b-5-b-c+1

           =a-c-4

    B+C=-b-c+1+b-c-4

           =-2c-3

\(\Rightarrow A+B\ne B+C\)

Đến đây thấy đề bài lỗi

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ấn vô link này ****

-(-a+b+c)+(b+c-1)

=a-b-c+b+c-1

=a-1   (1)

(b-c+6)-(7-a+b)+c

=b-c+6-7+a-b+c

=-1+a

=a-1   (2)

Từ (1)(2) => -(-a+b+c)+(b+c-1)=(b-c+6)-(7-a+b)+c

Ta có:

-(-a+b+c)+(b+c-1)

=a-b-c+b+c-1

=(b-b)+(c-c)+a-1

=0+0+a-1

=a-1

(b-c+6)-(7-a+b)+c

=b-c+6-7+a-b+c

=(b-b)+(c-c)+a+[(-7)+6]

=0+0+a-1

=a-1

Vì a-1=a-1

=>-(-a+b+c)+(b+c-1)=(b-c+6)-(7-a+b)-c

Ta có:

-(-a+b+c)+(b+c-1)

=a-b-c+b+c-1

=(b-b)+(c-c)+a-1

=0+0+a-1

=a-1

(b-c+6)-(7-a+b)+c

=b-c+6-7+a-b+c

=(b-b)+(c-c)+a+[(-7)+6]

=0+0+a-1

=a-1

Vì a-1=a-1

=>-(-a+b+c)+(b+c-1)=(b-c+6)-(7-a+b)-c

thank nha bạm

=-a-b-c-1+b-c-a+c+1

=-c-2a

Xét vế trái:

\(-\left(a+b+c+1\right)+\left(b-c\right)-\left(a-c-1\right)\\ =-a-b-c-1+b-c-a+c+1\\ =-c-2a\)