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![](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)
Ta có:
\(\frac{m}{n}+2017=\frac{n}{m}+2017\Rightarrow\frac{m}{n}=\frac{n}{m}\Rightarrow m^2=n^2\)
TH1: \(m=n\)
\(\Rightarrow x=1+2017=2018\)
TH2: \(-m=n\)
\(\Rightarrow x=-1+2017=2016\)
Vậy \(\left[{}\begin{matrix}x=2018\\x=2016\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(m+n\ne0.\)
\(\Rightarrow m+n+2017\ne2017.\)
Có:
\(x=\frac{m}{n+2017}=\frac{n}{m+2017}=\frac{2017}{m+n}\) và \(m+n+2017\ne2017.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(x=\frac{m}{n+2017}=\frac{n}{m+2017}=\frac{2017}{m+n}\)
\(\Rightarrow x=\frac{m+n+2017}{n+2017+m+2017+m+n}\)
\(\Rightarrow x=\frac{m+n+2017}{2m+2n+4034}\)
\(\Rightarrow x=\frac{m+n+2017}{2.\left(m+n+2017\right)}\)
\(\Rightarrow x=\frac{1}{2}.\)
Vậy \(x=\frac{1}{2}.\)
Chúc bạn học tốt!
Các bạn giúp ạ : @Vũ Minh Tuấn , @Băng Băng 2k6 , @Phạm Lan Hương , và cô @Akai Haruma
![](https://rs.olm.vn/images/avt/0.png?1311)
A=\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+2}+..........+\frac{2018}{2017^2+2017}\)
>\(\frac{2018}{2017^2+2017}+\frac{2018}{2017^2+2017}+........+\frac{2018}{2017^2+2017}\)
\(=\frac{2018}{2017^2+2017}.2017=\frac{2018.2017}{2017\left(2017+1\right)}=1\) (1)
Lại có:A<\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+1}+.........+\frac{2018}{2017^2+1}\)
\(=\frac{2018}{2017^2+1}.2017=\frac{2018.2017}{2017^2+1}=\frac{2017.\left(2017+1\right)}{2017^2+1}\)
\(=\frac{2017^2+2017}{2017^2+1}=\frac{2017^2+1+2016}{2017^2+1}=1+\frac{2016}{2017^2+1}< 2\) (2)
Từ (1) và (2) suy ra:1 < A < 2
Vậy A không phải là số nguyên
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^{2017}}{b^{2017}}=\frac{c^{2017}}{d^{2017}}\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)\(\Rightarrow\frac{a^{2017}}{b^{2017}}=\frac{c^{2017}}{d^{2017}}=\left(\frac{a+c}{b+d}\right)^{2017}\left(1\right)\)
\(\frac{a^{2017}}{b^{2017}}=\frac{c^{2017}}{d^{2017}}=\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\left(\frac{a+c}{b+d}\right)^{2017}\left(đpcm\right)\)
Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=b.k,c=d.k\)
Ta có:
\(\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\frac{\left(b.k\right)^{2017}+\left(d.k\right)^{2017}}{b^{2017}+d^{2017}}=\frac{b^{2017}.k^{2017}+d^{2017}.k^{2017}}{b^{2017}+d^{2017}}=\frac{k^{2017}.\left(b^{2017}+d^{2017}\right)}{b^{2017}+d^{2017}}=k^{2017}\) (1)
\(\left(\frac{a+c}{b+d}\right)^{2017}=\left(\frac{b.k+d.k}{b+d}\right)^{2017}=\left[\frac{k.\left(b+d\right)}{b+d}\right]^{2017}=k^{2017}\) (2)
Từ (1) và (2) suy ra \(\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\left(\frac{a+c}{b+d}\right)^{2017}\)
Vậy \(\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\left(\frac{a+c}{b+d}\right)^{2017}\)
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)