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 \(x=\frac{1}{2}.\)

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Vì \(\frac{n}{m+2017}=\frac{2017}{m+n}\Rightarrow n\left(m+n\right)=2017\left(m+2017\right)\Rightarrow n=2017\)

   \(\frac{m}{n+2017}=\frac{2017}{m+n}\Rightarrow2017\left(n+2017\right)=m\left(m+n\right)\Rightarrow m=2017\)

\(\Rightarrow x=\frac{2017}{2017+2017}=\frac{2017}{2017+2017}=\frac{2017}{2017+2017}=\frac{1}{2}\)

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 \(\left[{}\begin{matrix}x=2018\\x=2016\end{matrix}\right.\)

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

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

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

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

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

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)

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)

Theo bài ra ta có x + y + z = 2017

⇔\(\left\{{}\begin{matrix}2017-z=x+y\\2017-y=x+z\\2017-x=y+z\end{matrix}\right.\) (1)

Thay (1) vào \(A=\frac{x}{2017-z}+\frac{y}{2017-x}+\frac{z}{2017-y}\) ta được

\(A=\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{x+z}\)

Lại có \(\left\{{}\begin{matrix}x< x+y< x+y+z\\y< y+z< x+y+z\\z< x+z< x+y+z\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{x+y+z}< \frac{x}{x+y}< 1\\\frac{y}{x+y+z}< \frac{y}{y+z}< 1\\\frac{z}{x+y+z}< \frac{z}{x+z}< 1\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}\frac{x}{x+y+z}< \frac{x}{x+y}< \frac{x+z}{x+y+z}\\\frac{y}{x+y+z}< \frac{y}{y+x}< \frac{y+z}{x+y+z}\\\frac{z}{x+y+z}< \frac{z}{z+x}< \frac{z+y}{x+y+z}\end{matrix}\right.\)

( Áp dụng tính chất \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+c}{b+c}\) )

⇔ \(\frac{x+y+z}{x+y+z}< \frac{x}{x+y}+\frac{y}{y+x}+\frac{z}{x+z}< \frac{2.\left(x+y+z\right)}{x+y+z}\)

⇔ 1 < A < 2

⇔ A ko phải là số nguyên

Học tốt ~~ lâu hơn lm hình r c ạ ))

N = \(\frac{2016+2017}{2017+2018}=\frac{2016}{2017+2018}+\frac{2017}{2017+2018}\)

Ta có: \(\frac{2016}{2017}>\frac{2016}{2017+2018}\)

\(\frac{2017}{2016}>\frac{2017}{2017+2018}\)

Nên M > N

Ta thấy : \(\frac{2016+2017}{2017+2018}\)=\(\frac{2016}{2017+2018}\)+\(\frac{2017}{2017+2018}\)

Vì : \(\frac{2016}{2017}\)>\(\frac{2016}{2017+2018}\)

\(\frac{2017}{2018}\)>\(\frac{2017}{2017+2018}\)

Cộng vế với vế ta được : \(\frac{2016}{2017}\)+\(\frac{2017}{2018}\)> \(\frac{2016}{2017+2018}\)+\(\frac{2017}{2017+2018}\)

Hay M > N

Vậy M > N

Chúc bạn hok tốt !!