Question

cho các số a,b,c thỏa \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c};\)(a,b,c khác 0)

Tính \(N=\left(a^{15}+b^{15}\right)\left(b^{17}+c^{27}\right)\left(c^{2015}+a^{2015}\right)\)

Answer 1

Từ gt , ta có :

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

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

\(\Leftrightarrow\left(a+b\right)c\left(a+b+c\right)=-\left(a+b\right)ab\)

\(\Rightarrow0=\left(a+b\right)\left(ca+cb+c^2\right)-\left[-\left(a+b\right)ab\right]=\left(a+b\right)\left(ca+cb+c^2+ab\right)=\left(a+b\right)\left(c+a\right)\left(c+b\right)\)

\(\Rightarrow a+b=0\) hoặc \(c+a=0\) . Gỉa sử \(a=-b\) thì \(a^{15}=-b^{15}\) nên \(a^{15}+b^{15}=0\)

\(\Rightarrow N=0\)