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: \(\frac{1}{M^{2017}}+\frac{1}{N^{2017}}+\frac{1}{P^{2017}}=\frac{1}{M^{2017}+N^{2017}+P^{2017}}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì \(\frac{n}{m+2017}=\frac{2017}{m+n}\Rightarrow n\left(m+n\right)=2017\left(m+2017\right)\Rightarrow n=2017\)
\(\frac{m}{n+2017}=\frac{2017}{m+n}\Rightarrow2017\left(n+2017\right)=m\left(m+n\right)\Rightarrow m=2017\)
\(\Rightarrow x=\frac{2017}{2017+2017}=\frac{2017}{2017+2017}=\frac{2017}{2017+2017}=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
*Nếu \(m+n+2017\ne0\)thì theo t/c dãy tỉ số bằng nhau, ta được:
\(x=\frac{m}{n+2017}=\frac{n}{n+2017}=\frac{2017}{m+n}=\frac{1}{2}\)
*Nếu \(m+n+2017=0\)thì \(\hept{\begin{cases}m+n=-2017\\m+2017=-n\\n+2017=-m\end{cases}}\)
\(\Rightarrow x=\frac{m}{-m}=\frac{n}{-n}=\frac{2017}{-2017}=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
co m/n =2017/2017 => m/n=1 =>m=n => m+2017=n+2017
suy ra m+2017/n+2017 =1
ma m/n=1 => m/n=m+2017/n+2017
Ta có :
\(\frac{m}{n}=\frac{2017}{2017}\Leftrightarrow m=n\)
=> \(\frac{m+2017}{n+2017}=\frac{m+2017}{m+2017}=1=\frac{m}{n}\)
=> \(\frac{m}{n}=\frac{m+2017}{n+2017}\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Thay a+b+c=2017 vào \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2017}\) ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
\(\Rightarrow\frac{a+b}{ab}+\frac{a+b+c-c}{c\left(a+b+c\right)}=0\)\(\Rightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)
\(\Rightarrow\left(a+b\right)\left(\frac{1}{ab}+\frac{1}{c\left(a+b+c\right)}\right)=0\)\(\Rightarrow\left(a+b\right)\left(\frac{c\left(a+b+c\right)+ab}{abc\left(a+b+c\right)}\right)=0\)
\(\Rightarrow\left(a+b\right)\left(\frac{c\left(b+c\right)+ca+ab}{abc\left(a+b+c\right)}\right)=0\)
\(\Rightarrow\left(a+b\right)\left[c\left(b+c\right)+ca+ab\right]=0\)
\(\Rightarrow\left(a+b\right)\left[c\left(b+c\right)+a\left(b+c\right)\right]=0\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Rightarrow\)\(a+b=0\) hoặc \(b+c=0\) hoặc \(c+a=0\)
\(\Rightarrow\)\(c=2017\)hoặc \(a=2017\) hoặc \(b=2017\left(đpcm\right)\)
\(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}-\frac{1}{m+n+p}=0\)
\(\Leftrightarrow\frac{m+n}{mn}+\frac{m+n}{p\left(m+n+p\right)}=0\)
\(\Leftrightarrow\left(m+n\right)\left(\frac{pm+pn+p^2+mn}{mnp\left(m+n+p\right)}\right)=0\)
\(\Leftrightarrow\left(m+n\right)\left(n+p\right)\left(p+m\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}m=-n\\m=-p\\p=-n\end{matrix}\right.\)
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)