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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
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: \(=2\left(a-b\right)\left(c-b\right)-2\left(a-b\right)\left(c-a\right)+2\left(b-c\right)\left(a-c\right)\)
\(=2\left(a-b\right)\left(c-b-c+a\right)+2\left(b-c\right)\left(a-c\right)\)
\(=2\left(a-b\right)\left(a-b\right)+2\left(b-c\right)\left(a-c\right)\)
\(=2\left(a^2-2ab+b^2+ab-bc-ac+c^2\right)\)
\(=2\left(a^2+b^2-ab-bc-ac+c^2\right)\)
\(=\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\)
a)a(b2+c2)+b(a2+c2)+c(a2+b2)+2abc
=ab2+ac2+ba2+bc2+ca2+cb2+2abc
=(ab2+ba2)+(ac2+bc2)+(ca2+abc)+(cb2+abc)
=ab(a+b)+c2(a+b)+ca(a+b)+cb(a+b)
=(a+b)(ab+c2+ca+cb)
=(a+b)(a+c)(b+c)
b)a3-b3-c3-3abc
=(a-b)3-c3+3ab(a-b)-3abc
=(a-b-c)[(a-b)2+(a-b)c+c2]+3ab(a-b-c)
=(a-b-c)(a2-2ab+b2+ac-bc+c2+3ab)
=(a-b-c)(a2+b2+c2+ab-bc+ca)