Áp dụng tính chất dãy tỉ số bằng nhau ta có: 

\(\frac{x+y+2015}{z}=\frac{y+z-2016}{x}=\frac{z+x+1}{y}.\)

\(=\frac{x+y+2015+y+z-2016+z+x+1}{x+y+z}\)\(=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

Do đó x+y+z=1 => x+y=1-z => \(\frac{2016-z}{z}=2\Rightarrow2016-z=2z\Leftrightarrow2016=3z\)

=> z= 672

Tương tự : x= -2015/3; y=2/3

x=2015/3

y=2/3

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)

\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)

\(\Rightarrow x=165;y=20;z=25\)

x=by+cz;y=ax+cz;z=ax+by

=>x+y+z=2(ax+by+cz)

\(\Leftrightarrow\frac{x+y+z}{2}=ax+by+cz\)

\(\Leftrightarrow y+z=\frac{x+y+z}{2}+ax;z+x=\frac{x+y+z}{2}+by;x+y=\frac{x+y+z}{2}+cz\)

\(\Leftrightarrow\frac{y+z-x}{2}=ax;\frac{z+x-y}{2}=by;\frac{x+y-z}{2}=cz\)

\(\Leftrightarrow\frac{y+z-x}{2x}=a;\frac{z+x-y}{2y}=b;\frac{x+y-z}{2z}=c\)

\(\Rightarrow A=\frac{1}{1+\frac{x+y-z}{2z}}+\frac{1}{1+\frac{y+z-x}{2x}}+\frac{1}{1+\frac{z+x-y}{2y}}=\frac{1}{\frac{x+y+z}{2x}}+\frac{1}{\frac{x+y+z}{2y}}+\frac{1}{\frac{x+y+z}{2z}}\)

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

thiếu đề

Ta có : 

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

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

\(z+x=\frac{1}{4}\)

\(\Rightarrow\)\(x+y+y+z+z+x=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)

\(\Rightarrow\)\(2x+2y+2z=\frac{13}{12}\)

\(\Rightarrow\)\(2\left(x+y+z\right)=\frac{13}{12}\)

\(\Rightarrow\)\(x+y+z=\frac{13}{12}:2\)

\(\Rightarrow\)\(x+y+z=\frac{13}{24}\)

Do đó : 

\(x+y+z=\frac{13}{24}\)

\(\Rightarrow\)\(x=\frac{13}{24}-\left(y+z\right)=\frac{13}{24}-\frac{1}{3}=\frac{5}{24}\)

\(\Rightarrow\)\(y=\frac{13}{24}-\left(z+x\right)=\frac{13}{24}-\frac{1}{4}=\frac{7}{24}\)

\(\Rightarrow\)\(z=\frac{13}{24}-\left(x+y\right)=\frac{13}{24}-\frac{1}{2}=\frac{1}{24}\)

Vậy \(x=\frac{5}{24};y=\frac{7}{24};z=\frac{1}{24}\)

Chúc bạn học tốt ~

Ta có:

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

\(=1+\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}=3+\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\)

\(\Rightarrow x+y+z=\frac{3+\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)}{\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}}=\frac{3+\frac{7}{10}}{\frac{2}{5}}=\frac{37}{4}\)

Ta có :

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

\(=1+\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}=3+\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\)

\(\Rightarrow x+y+z=\frac{3+\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)}{\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}}=\frac{3+\frac{7}{10}}{\frac{2}{5}}=\frac{37}{4}\)

Ta có:

\(x+y+y+z+z+x=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)

\(\Rightarrow2\left(x+y+z\right)=\frac{13}{12}\Leftrightarrow x+y+z=\frac{13}{12}.\frac{1}{2}=\frac{13}{24}\)

\(\cdot x+y=\frac{1}{2}\Leftrightarrow z=\frac{13}{24}-\frac{1}{2}=\frac{1}{24}\)

\(\cdot y+z=\frac{1}{3}\Leftrightarrow x=\frac{13}{24}-\frac{1}{3}=\frac{5}{24}\)

\(\cdot y=\frac{13}{24}-\frac{1}{24}-\frac{5}{24}=\frac{7}{24}\)

Vậy \(x=\frac{5}{24};y=\frac{7}{24};z=\frac{1}{24}\)

Ta có: y + z = \(\frac{1}{3}\); z + x = \(\frac{1}{4}\).

=> y lớn hơn x : \(\frac{1}{3}-\frac{1}{4}=\frac{1}{12}\).

x + y = \(\frac{1}{2}\)và  y - x = \(\frac{1}{12}\)=> x = \(\left(\frac{1}{2}-\frac{1}{12}\right):2=\frac{5}{24}\)

=> y = \(\frac{1}{2}-\frac{5}{24}=\frac{7}{24}\)

=> z = \(\frac{1}{4}-\frac{5}{24}=\frac{1}{24}\)