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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
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: \(=2\left(a-b\right)\left(c-b\right)-2\left(a-b\right)\left(c-a\right)+2\left(b-c\right)\left(a-c\right)\)
\(=2\left(a-b\right)\left(c-b-c+a\right)+2\left(b-c\right)\left(a-c\right)\)
\(=2\left(a-b\right)\left(a-b\right)+2\left(b-c\right)\left(a-c\right)\)
\(=2\left(a^2-2ab+b^2+ab-bc-ac+c^2\right)\)
\(=2\left(a^2+b^2-ab-bc-ac+c^2\right)\)
\(=\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\)
a) Đặt: x = a- b; y = b - c ; z = c- a
Ta có: x + y + z = 0
=> \(A=x^3+y^3+z^3=3xyz+\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=3xyz\)
=> \(A=3xyz=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
b) Đặt: \(a=x^2-2x\)
Ta có: \(B=a\left(a-1\right)-6=a^2-a-6=\left(a+2\right)\left(a-3\right)=\left(x^2-2x+2\right)\left(x^2-2x-3\right)\)
\(=\left(x^2-2x+2\right)\left(x+1\right)\left(x-3\right)\)
d) \(D=4\left(x^2+2x-8\right)\left(x^2+7x-8\right)+25x^2\)
Đặt: \(x^2-8=t\)
Ta có: \(D=4\left(t+2x\right)\left(t+7x\right)+25x^2\)
\(=4t^2+36xt+81x^2=\left(2t+9x\right)^2\)
\(=\left(2x^2+9x-16\right)^2\)
Ta có:\(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(a^2+b^2\right).c^2+\left(a^2+b^2\right).d^2\)
\(=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)
\(=a^2c^2-2abcd+b^2d^2+b^2c^2+2abcd+a^2d^2\)
\(=\left(ac-bd\right)^2+\left(bc+ad\right)^2\) là tổng 2 bình phương
\(\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)\(=\left(a^2c^2+b^2d^2+2abcd\right)+\left(a^2d^2+b^2c^2-2abcd\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)