Lời giải:

Ta có:

\(A=\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{zx+z+1}\)

\(A=\frac{xz}{xyz+xz+z}+\frac{y.xz}{yz.xz+y.xz+xz}+\frac{z}{zx+z+1}\)

\(A=\frac{xz}{1+xz+z}+\frac{1}{z+1+xz}+\frac{z}{xz+z+1}\) (thay \(xyz=1\) )

\(A=\frac{xz+1+z}{1+xz+z}=1\)

a) Cộng cả 3 đẳng thức trên ta có:

2(x + y + z) = 1/2 +1/3 + 1/4 = 13/12 => x + y + z = 13/24 (*)

z = 13/24 - 1/2 = 1/24

x = 13/24 - 1/3 = 5/24

y = 13/24 - 1/4 = 7/24.

b) Nhân cả 3 đẳng thức ta có: x²y²z² = 1/16 => xyz = 1/4 hoặc -1/4

ko hiểu đề cho lắm

a)Ta có:

\(\left\{{}\begin{matrix}x+y=\frac{1}{3}\\y+z=\frac{-1}{4}\\z+x=\frac{1}{5}\end{matrix}\right.\)

\(\Rightarrow\left(x+y\right)+\left(y+z\right)+\left(z+x\right)=\frac{1}{3}+\frac{-1}{4}+\frac{1}{5}\)

\(\Rightarrow2\left(x+y+z\right)=\frac{17}{60}\)

\(\Rightarrow x+y+z=\frac{17}{60}:2=\frac{17}{120}\)

\(\Rightarrow\left\{{}\begin{matrix}z=\frac{-23}{120}\\x=\frac{47}{120}\\y=\frac{-7}{120}\end{matrix}\right.\)

b)Ta có:

\(\left\{{}\begin{matrix}xy=\frac{3}{5}\\yz=\frac{4}{5}\\zx=\frac{3}{4}\end{matrix}\right.\)

\(\Rightarrow xyyzzx=\frac{3}{5}.\frac{4}{5}.\frac{3}{4}=\frac{9}{25}\)

\(\Rightarrow\left(xyz\right)^2=\frac{9}{25}\Rightarrow\left[{}\begin{matrix}xyz=\frac{3}{5}\\xyz=-\frac{3}{5}\end{matrix}\right.\)

TH1: \(xyz=\frac{3}{5}\)

\(\Rightarrow\left\{{}\begin{matrix}z=1\\x=\frac{3}{4}\\y=\frac{4}{5}\end{matrix}\right.\)

TH2:

\(xyz=-\frac{3}{5}\)

\(\Rightarrow\left\{{}\begin{matrix}z=-1\\x=-\frac{3}{4}\\y=-\frac{4}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y+z\right)^2=81\\xy+yz+xz=27\\\dfrac{xy+xz+zy}{xyz}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+z^2+2\left(xy+yz+xz\right)=81\\xy+yz+xz=27\\xyz=27\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+z^2=27\\xy+yz+xz=27\\xyz=27\end{matrix}\right.\Leftrightarrow x^2+y^2+z^2=xy+yz+xz=xyz\)

theo bđt ta có \(x^2+y^2+z^2\ge xy+xz+yz\)

để \(x^2+y^2+z^2=xy+xz+yz\) khi \(x=y=z=3\)

Đặt \(k=\frac{x}{8}=\frac{y}{3}=\frac{z}{10}\)

Ta có: \(x=8k;y=3k;z=10k\)  (*)

Thay vào đẳng thức \(xy+yz+zx=206\) ta được:

  \(8k.3k+3k.10k+10k.8k=206\)

\(\Leftrightarrow24k^2+30k^2+80k^2=206\)

\(\Leftrightarrow24k^2+30k^2+80k^2=206\)

\(\Rightarrow k=\pm\sqrt{\frac{103}{67}}\)

Thay k vào (*) tính được x, y, z