\(\left(-a+b+c\right)+\left(b+c-1\right)=-a+b+c+b+c-1=2b+2c-a-1\)
\(\left(b-c+6\right)-\left(7-a+b\right)+c=b-c+6-7+a-b+c=b-b+c-c+a+6-7=a-1\)
Không bằng nhau
(-a+b+c)+(b+c-1)=(b-c+6)-(7-a+b)+c
-a+b+c+b+c-1=b-c+6-7+a-b+c
-a+2b+2c-1=-1+a
2b+2c=-1+1+a+a
2(b+c)=2a
b+c=a
để (-a+b+c)+(b+c-1)=(b-c+6)-(7-a+b)+c thì b+c=a