Question

Cho M, N, P là các số khác 0và M+N+P\(\ne\)0 thỏa mãn \(\frac{1}{M}+\frac{1}{N}+\frac{1}{P}=\frac{1}{M+N+P}\). Chứng minh: \(\frac{1}{M^{2017}}+\frac{1}{N^{2017}}+\frac{1}{P^{2017}}=\frac{1}{M^{2017}+N^{2017}+P^{2017}}\)

Answer 1

\(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}-\frac{1}{m+n+p}=0\)

\(\Leftrightarrow\frac{m+n}{mn}+\frac{m+n}{p\left(m+n+p\right)}=0\)

\(\Leftrightarrow\left(m+n\right)\left(\frac{pm+pn+p^2+mn}{mnp\left(m+n+p\right)}\right)=0\)

\(\Leftrightarrow\left(m+n\right)\left(n+p\right)\left(p+m\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}m=-n\\m=-p\\p=-n\end{matrix}\right.\)

Cả 3 TH là như nhau

Ví dụ như TH1: \(\frac{1}{m^{2017}}+\frac{1}{-m^{2017}}+\frac{1}{p^{2017}}=\frac{1}{p^{2017}}\)

\(\frac{1}{m^{2017}-m^{2017}+p^{2017}}=\frac{1}{p^{2017}}\) (đpcm)