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Ta có a+b+c=0 => \(a+b=-c\Rightarrow\left(a+b\right)^3=-c^3\Rightarrow a^3+b^3+c^3=-3ab\left(a+b\right)=3ab\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow ab+bc+ca=0\)
\(a^6+b^6+c^6=\left(a^3\right)^2+\left(b^3\right)^2+\left(c^3\right)^2=\left(a^3+b^3+c^3\right)^2-2\left(a^3b^3+b^3c^3+c^3a^3\right)\)
\(ab+bc+ca=0\Rightarrow a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2\)
Do đó: \(a^6+b^6+c^6=\left(3abc\right)^2-2\cdot3a^2b^2c^2=3a^2b^2c^2\)
Vậy \(\frac{a^6+b^6+c^6}{a^3+b^3+c^3}=\frac{3a^2b^2c^2}{3abc}=abc\left(đpcm\right)\)
Bạn tham khảo :
Ta có :
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)
\(\Rightarrow\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+3=1\)
\(\Rightarrow\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+2=0\)
\(\Rightarrow abc\left(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+2\right)=abc.0\)
\(\Rightarrow a^2b+b^2c+a^2c+b^2a+c^2a+c^2b+2abc=0\)
\(\Rightarrow\left(a^2b+ab^2\right)+\left(b^2c+abc\right)+\left(a^2c+abc\right)+\left(c^2a+c^2b\right)=0\)
\(\Rightarrow ab\left(a+b\right)+bc\left(a+b\right)+ac\left(a+b\right)+c^2\left(a+b\right)=0\)
\(\Rightarrow\left(ab+bc+ac+c^2\right)\left(a+b\right)=0\)
\(\Rightarrow\left[\left(ab+bc\right)+\left(ac+c^2\right)\right]\left(a+b\right)=0\)
\(\Rightarrow\left[b\left(a+c\right)+c\left(a+c\right)\right]\left(a+b\right)=0\)
\(\Rightarrow\left(a+c\right)\left(b+c\right)\left(a+b\right)=0\)
TH1 : \(a+c=0\)
\(\Rightarrow a=-c\)
\(\Rightarrow c^{2006}=a^{2006}\)
\(\Rightarrow P=\left(a^{2004}-b^{2004}\right)\left(b^{2005}+c^{2005}\right)\left(c^{2006}-a^{2006}\right)\)
\(=\left(a^{2004}-b^{2004}\right)\left(b^{2005}+c^{2005}\right)0\)
\(=0\)
CMTT đều có \(P=0\)
Vậy ...
từ giả thiết ta có
a+b+c=0
<=> a=-(b+c0
a2=b2 +c2 +2bc
tương tự b2=a2+c2+2ac
c2=a2+b2+2ab
thay vào Q ta đc
\(Q=\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+c^2-b^2}\)
\(Q=\frac{1}{a^2+b^2-a^2-b^2-2ab}+\frac{1}{b^2+c^2-b^2-c^2-2bc}+\frac{1}{a^2+c^2-a^2-c^2-2ac}\)
\(Q=\frac{-1}{2ab}-\frac{1}{2bc}-\frac{1}{2ac}\)
\(Q=\frac{-b-a-c}{2abc}\)
\(Q=\frac{-\left(a+b+c\right)}{2abc}\)
\(Q=0\)
Vậy với a,b,c khác 0, a+b+c=0 thì Q=0