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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
\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
1. \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left[\left(abc\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2+c^2-ac-bc\right)-3ab\left(a+b+c\right)\)
\(\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc+2ab-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
2. \(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
3.Còn có a + b + c = 0 nữa mà bn.
\(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-ac-bc=0\end{matrix}\right.\)
+ \(a^2+b^2+c^2-ab-bc-ac=0\)
\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\ \left(c-a\right)^2=0\end{matrix}\right.\)
\(\Rightarrow a=b=c\)
bài 1: (A+B)3=A3+3A2B+3AB2+B3 (1)
A3+B3=(A+B)(A2+AB+B2) (2)
Tính A3+B3+C3=?
Bài 2: tính giá trị biểu thức
M= yz/x2 + xz/y2 +xy/z2 với xy+yz+xz=0 và xyz \(^{_{ }\ne}\) 0