Question

cho a+b+C ≠0 , \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

Tính A=\(\left(a^{2011}+b^{2011}+c^{2011}\right)\left(\dfrac{1}{a^{2011}}+\dfrac{1}{b^{2011}}+\dfrac{1}{c^{2011}}\right)\)

Answer 1

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a+b+c}=0\)

\(\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\)

\(\left(a+b\right)\left(\dfrac{1}{ab}+\dfrac{1}{c\left(a+b+c\right)}\right)=0\)

\(\left(a+b\right)\left(\dfrac{ab+bc+ca+c^2}{abc\left(a+b+c\right)}\right)=0\)

\(\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)\(\Rightarrow A=1\)