a. ( a + b + c)² + a² + b² + c²

= a² + b² + c² + 2ab + 2ac + 2bc + a² + b² + c²

= (a+b)² + (b+c)² + (a+c)²

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

đặt a - b = x; b-c = y; c-a = z => x + y + z = 0 (1)

ta có: 2.x.(-y) + 2.(-x).z + 2.y.(-z)

= -2xy - 2xz - 2yz  = -2.(xy+xz+yz)  

ta có: (x+y+z)² = x² + y² + z² + 2xy + 2yz + 2xz

 0² = x² + y² + z² + 2.(xy+yz+xz)

=> x² + y² + z²  = -2.(xy+yz+xz) (2)

Từ (2) => 2.(a-b).(c-b) + 2.(b-a) .(c-a) + 2.(b-c).(a-c) = x² + y² + z²

= (a-b)² + (b-c)² + (c-a)²

1a) a² + b² + c² + 2ab + 2bc + 2ca + a² + b² + c²

= ( a² + 2ab +b² ) + ( a² + 2ac + c² ) + ( b² + 2bc + c² )

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

1b) 2.( ac - ab - bc + b² ) + 2.( bc - ba - ac + a² ) + 2.( ba - bc - ca + c² )

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

= 2a² + 2b² + 2c² - 2ab - 2ac - 2bc

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

= (a-b)² + (a-c)² + (b-c)²

Bài 2 :

a ) \(A=\left(a+b+c\right)^2+a^2+b^2+c^2\)

\(A=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\)

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

\(A=\left(a+b\right)^2+\left(a+c\right)^2+\left(b+c\right)^2\)

a: \(=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2\)

\(=a^2+2ab+b^2+b^2+2bc+c^2+a^2+2ac+c^2\)

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

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

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

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

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

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

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

bố éo biết

\(25{x^2} + 20xy + 4{y^2} = {\left( {5x} \right)^2} + 2.5x.2y + {\left( {2y} \right)^2} = {\left( {5x + 2y} \right)^2}\)

Chọn D.