(a+b-c)^2 + (a-b+c)^2 - 2(b-c)^2 

= (a + b - c)^2 - (b - c)^2 + (a - b + c)^2 - (b - c)^2 

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

= a^2 + a^2 

= 2.a^2

\(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2bc\)

\(=\left(a-c\right)^2+2b\left(a-c\right)+b^2-\left(a-c\right)^2-2ab+2bc\)

\(=2b\left(a-c\right)+b^2-2ab+2bc\)

\(=2ab-2bc+b^2-2ab+2bc=b^2\)

  (a + b + c)^2 + (a - b - c)^2 +( b - c - a) ^2 + (c - a - b)^2 

= (a + b + c)^2 + (a + b - c)^2 + (a - b - c)^2 + (a - b + c)^2 

= (a + b)^2 + 2c(a + b) + c^2 + (a + b)^2 - 2c(a + b) + c^2 + 
(a - b)^2 - 2c(a - b) + c^2 + (a - b)^2 + 2c(a - b) +c^2 

= 2(a + b)^2 + 2c^2 + 2(a - b)^2 + 2c^2 

= 2[(a + b)^2 + (a - b)^2] + 4c^2 

=2(2a^2 + 2b^2) + 4c^2 

= 4(a^2 + b^2 + c^2)

Ta có  : (a + b - c)² + (a - b + c)² - 2(b - c)²

= a² + b² + c² + 2ab - 2bc - 2ca + a² + b² + c² + 2ca - 2ab - 2bc - 2(b² - 2bc + c²)

= a² + b² + c² + 2ab - 2bc - 2ca + a² + b² + c² + 2ca - 2ab - 2bc - b² + 2bc - c²

= 2a² + b² + c² - 2bc 

Ta có A=(-a-b+c)-(-a-b-c)

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

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

=0

Khi a=1 ; b=-1;c=2

Thì A =(-1+1+2)-(-1+1-2)

=2-2

=0

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

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

= 0 + ( b-b ) + ( c+c )

=0+0+  2x

=2x

(-a-b+c)-(-a-b-c)=-a-b+c+a+b+c=2c

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

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

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

=0+0+c.2

=c2

= -a - b + c + a + b +c -2b

= 2c- 2b

=2. ( c - b)

hok tốt

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

\(=-a-b+c+a+b+c-2b\)

\(=2c-2b=2.\left(c-b\right)\)