Question

Cho các số a,b,c thỏa mãn \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\left(a,b,c\ne0\right)\).

Tính giá trị của biểu thức \(N=\left(a^{15}+b^{15}\right)\left(b^{27}+c^{27}\right)\left(c^{2015}+a^{2015}\right)\)

Answer 1

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

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

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

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

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

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

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

\(\Rightarrow N=0\)