Đặt : \(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=M\)

\(\Rightarrow\left(x+y+z\right).M=\frac{1}{672}.2017\)

\(\Rightarrow1+\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}=\frac{2016}{672}+\frac{1}{672}\)

\(\Rightarrow3+\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=3+\frac{1}{672}\)

\(\Rightarrow\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=\frac{1}{672}\)

Nhân cả 2 vế với \(x+y+z\),ta được:

\(\left(x+y+z\right)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{1}{672}\cdot2017\)

\(\Rightarrow\frac{x+y+z}{x+y}+\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}=\frac{2017}{672}\)

\(\Rightarrow3+\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\frac{2017}{672}\)

\(\Rightarrow C=\frac{1}{672}\)

Nhanh k cho nè

làm lần lượt nhá,dài dòng quá khó coi.ahihihi!

\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{7\left(\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)

\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}=\frac{1}{4}\)

ta có: \(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}=\frac{1}{90}.\)

\(\Rightarrow2007.\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)=2007\cdot\frac{1}{90}\)

\(\frac{2007}{x+y}+\frac{2007}{y+z}+\frac{2007}{x+z}=\frac{223}{10}\)

mà x+y+z = 2007

\(\Rightarrow\frac{x+y+z}{x+y}+\frac{x+y+z}{y+z}+\frac{x+y+z}{x+z}=\frac{223}{10}\)

\(1+\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{x+z}=\frac{223}{10}\)

\(\Rightarrow\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{x+z}=\frac{223}{10}-3=\frac{193}{10}\)

Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)  ( x, y , z khác 0 )  (@)

<=> \(\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)

<=> \(\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)

<=> x + y = 0  (1) 

hoặc: \(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}=0\)(2)

(2) <=> \(zx+zy+z^2+xy=0\)

<=> \(z\left(x+z\right)+y\left(x+z\right)=0\)

<=> \(\left(x+z\right)\left(y+z\right)=0\)

<=> x + z = 0 hoặc y + z = 0 

<=> x = - z hoặc y = -z 

(1) <=> x = - y 

Vậy: (@) <=> x = - y hoặc y = -z hoặc z = - x

Vì vị trí của x, y, z có vai trò như nhau. G/S: x = - y

khi đó: \(\frac{1}{x^{2017}}+\frac{1}{y^{2017}}+\frac{1}{z^{2017}}=\frac{1}{\left(-y\right)^{2017}}+\frac{1}{y^{2017}}+\frac{1}{z^{2017}}=\frac{1}{z^{2017}}\)

và: \(\frac{1}{x^{2017}+y^{2017}+z^{2017}}=\frac{1}{z^{2017}}\)

Do vậy: \(\frac{1}{x^{2017}}+\frac{1}{y^{2017}}+\frac{1}{z^{2017}}=\)\(\frac{1}{x^{2017}+y^{2017}+z^{2017}}\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)(vì x + y + z khác 0)

=> \(\frac{1}{x+y+z}=2\) => x + y + z = 1/2

=> \(\hept{\begin{cases}\frac{y+z+1}{x}=2\\\frac{x+z+2}{y}=2\\\frac{x+y-3}{z}=2\end{cases}}\) => \(\hept{\begin{cases}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\end{cases}}\) => \(\hept{\begin{cases}3x=x+y+z+1\\3y=x+y+z+2\\3z=x+y+z-3\end{cases}}\)=> \(\hept{\begin{cases}3x=\frac{3}{2}\\3y=\frac{5}{2}\\3z=-\frac{5}{2}\end{cases}}\)=> \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{5}{6}\\z=-\frac{5}{6}\end{cases}}\)

Khi đó: A = \(2016\cdot\frac{1}{2}+\left(\frac{5}{6}\right)^{2017}-\left(\frac{5}{6}\right)^{2017}=1008\)

Ta có \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}\)

                                                                                                                 \(=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

Khi đó \(\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\)

Lại có \(\frac{y+z+1}{x}=2\Rightarrow y+z+1=2x\Rightarrow x+y+z+1=3x\Rightarrow\frac{1}{2}+1=3x\Rightarrow3x=\frac{3}{2}\)

=> x = 1/2 

Lại có \(\frac{x+z+2}{y}=2\Rightarrow x+z+2=2y\Rightarrow x+y+z+2=3y\Rightarrow\frac{1}{2}+2=3y\Rightarrow3y=\frac{5}{2}\)

=> y = 5/6

Lại có x + y + z = 1/2

=> 1/2 + 5/6 + z = 1/2

=> 5/6 + z = 0

=> z = -5/6

Khi đó A = 2016X + y²⁰¹⁷ + z²⁰¹⁷

= 2016.1/2 + (5/6)²⁰¹⁷ - (5/6)²⁰¹⁷

= 1008

Vậy A = 1008