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Ta có:
a3 + b3 + c3 - 3abc
= (a + b)3 + c3 - 3ab(a + b) - 3abc
= (a + b + c)3 - 3(a + b)c(a + b + c) - 3ab(a + b + c)
= (a + b + c)[(a + b + c)2 - 3(a + b)c - 3ab]
= (a + b + c)(a2 + b2 + c2 + 2ab + 2bc + 2ac - 3ac - 3bc - 3ab)
= (a + b + c)(a2 + b2 + c2 - ab - bc - ac) = 3abc - 3abc = 0
=> a + b + c = 0 hay a2 + b2 + c2 - ab - bc - ac = 0
I => 2(a2 + b2 + c2 - ab - bc - ac) = 0
I => 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ac = 0
I => (a - b)2 + (b - c)2 + (a - c)2 = 0
I => a - b = 0 hay b - c = 0 hay a - c = 0
I => a = b I => b = c I => a = c
I => a = b = c
a + b + c = 0 => a + b = -c
=>(a + b)3 = (-c)3
=>a3 + b3 +3a2b + 3ab2 = (-c)3
=>a3 + b3 + c3 +3ab(a + b) = 0
=>a3 + b3 + c3 = 3abc
a)\(ab\left(a+b\right)-bc\left(b+c\right)+ac\left(a-c\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(a-c\right)\)
b)\((a+b)(a^2-b^2)+(b+c)(b^2-c^2)+(c+a)(c^2-a^2)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
c)\(a^2b^2(a-b)+b^2c^2(b-c)+c^2a^2(c-a)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(ab+bc+ca\right)\)
d)\(a^4(b-c)+b^4(c-a)+c^4(a-b)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a^2+b^2+c^2+ab+bc+ca\right)\)
Bài 2:
\(\left(a+b+c\right)^2=3\left(ab+bc+ac\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac-3ab-3ac-3bc=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
=>a=b=c
Cần CM :\(a^2+b^2+c^2-ab-bc-ca\)>=0
<=>\(2\cdot a^2+2\cdot b^2+2\cdot c^2-2ab-2bc-2ca\)>=0(1)
ta có \(2a^2+2b^2+2c^2-2ab-2bc-2ca\)=\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\)
=\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2>=0\) =>(1) (luôn đúng)
vậy suy ra đpcm
Dấu = khi a=b=c
Ta có ( a - b - c )2 >= 0
= ( a-b )2 - 2(a-b)c + c2 >= 0
= a2 - 2ab + b2 - 2ac + 2bc + c2 >= 0
= a2 + b2 + c2 - 2 ( ab - bc + ac ) >=0 (dpcm)
\(a^2\)+\(b^2\)+\(c^2\)-ab-bc-ca\(\ge\)0
<=> 2\(a^2\)+2\(b^2\)+2\(c^2\)-2ab-2bc-2ac\(\ge\)0
<=> (\(a^2\)-2ab+\(b^2\)) +(\(b^2\)-2bc+\(c^2\))+(\(c^2\)-2ca+\(a^2\))\(\ge\)0
<=> \(\left(a-b\right)^2\)+\(\left(b-c\right)^2\)+\(\left(c-a\right)^2\)\(\ge\)0
vì \(\left(a-b\right)^2\)\(\ge\)0
\(\left(b-c\right)^2\)\(\ge\)0
\(\left(c-a\right)^2\)\(\ge\)0
<=>\(\left(a-b\right)^2\)+\(\left(b-c\right)^2\)+\(\left(c-a\right)^2\)\(\ge\)0
vậy\(a^2\)+\(b^2\)+\(c^2\)-ab-bc-ca\(\ge\)0
dấu = xảy ra khi
a-b=0=>a=b
b-c=0=> b=c
c-a=0=> c=a
=> a=b=c
Tham khảo tại link sau : http://olm.vn/hoi-dap/question/721476.html
a2+b2+c2=ab+bc+ac
2a2+2b2+2c2=2ab+2bc+2ac
2a^2+2b^2+2c^2-2ab-2bc-2ac=0
(a-b)2+(a-c)2+(b-c)2=0
=> a=b=c
k co mình cái