Dễ thấy tập nghiệm \(\left(x;y;z\right)=\left(0;0;0\right)\) thỏa mãn.

Xét \(xyz\ne0\), hệ tương đương với :

\(\left\{{}\begin{matrix}\frac{x+y}{xy}=\frac{5}{6}\\\frac{y+z}{yz}=\frac{7}{12}\\\frac{x+z}{xz}=\frac{3}{4}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}=\frac{5}{6}\left(1\right)\\\frac{1}{y}+\frac{1}{z}=\frac{7}{12}\left(2\right)\\\frac{1}{x}+\frac{1}{z}=\frac{3}{4}\left(3\right)\end{matrix}\right.\)

Có \(2\cdot\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{5}{6}+\frac{7}{12}+\frac{3}{4}=\frac{13}{6}\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{13}{12}\)

+) Từ (1) suy ra \(\frac{1}{z}=\frac{13}{12}-\frac{5}{6}=\frac{1}{4}\Leftrightarrow z=4\)

+) Từ (2) suy ra \(\frac{1}{x}=\frac{13}{12}-\frac{7}{12}=\frac{1}{2}\Leftrightarrow x=2\)

+) Từ (3) suy ra \(\frac{1}{y}=\frac{13}{12}-\frac{3}{4}=\frac{1}{3}\Leftrightarrow y=3\)

Vậy tập nghiệm của hệ là \(\left(x;y;z\right)\in\left\{\left(0;0;0\right);\left(2;3;4\right)\right\}\)

Nhận thấy \(x=y=z=0\) là 1 nghiệm

Với \(x;y;z\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}=\frac{5}{12}\\\frac{1}{y}+\frac{1}{z}=\frac{5}{18}\\\frac{1}{z}+\frac{1}{x}=\frac{13}{36}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}=\frac{1}{4}\\\frac{1}{y}=\frac{1}{6}\\\frac{1}{z}=\frac{1}{9}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=4\\y=6\\z=9\end{matrix}\right.\)

Vậy hệ có 2 bộ nghiệm \(\left(x;y;z\right)=\left(0;0;0\right);\left(4;6;9\right)\)

b, Ta có : \(\left\{{}\begin{matrix}x+y+z=3\\y+z+t=4\\z+t+x=5\\t+x+y=6\end{matrix}\right.\)

=> \(x+y+z+y+z+t+z+t+x+t+x+y=18\)

=> \(3\left(x+y+z+t\right)=18\)

=> \(x+y+z+t=6\)

=> \(x+y+z+t=x+y+t\)

=> \(z=0\)

=> \(\left\{{}\begin{matrix}x+y=3\\y+t=4\\x+t=5\\x+y+t=6\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x+y=3\\y+t=4\\x+t=5\\y+5=6\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x+1=3\\t+1=4\\x+t=5\\y=1\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=2\\t=3\\x+t=5\\y=1\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}x=2\\y=1\\z=0\\t=3\end{matrix}\right.\)

a, Ta có : \(\left\{{}\begin{matrix}7xy=12\left(x+y\right)\\9yz=20\left(y+z\right)\\8zx=15\left(z+x\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}7xy-12x-12y=0\\9yz-20y-20z=0\\8zx-15z-15x=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=\frac{12y}{7y-12}\\y=\frac{20z}{9z-20}\\x=\frac{15z}{8z-15}\end{matrix}\right.\)

=> \(12y\left(8z-15\right)=15z\left(7y-12\right)\)

=> \(96yz-180y=105yz-180z\)

=> \(105yz-96yz=-180y+180z\)

=> \(9yz=-180y+180z\)

=> \(180z-180y=20y+20z\)

=> \(180z-20z=180y+20y=160z=200y\)

=> \(y=\frac{4}{5}z\)

=> \(\frac{20z}{9z-20}=\frac{4z}{5}\)

=> \(4z\left(9z-20\right)=100z\)

=> \(36z^2-180z=0\)

=> \(\left[{}\begin{matrix}z=5\\z=0\end{matrix}\right.\)

TH1 : z = 0 .

=> \(x=y=z=0\)

TH₂ : z = 5 .

=> \(\left\{{}\begin{matrix}7xy=12\left(x+y\right)\\45y=20\left(y+5\right)\\40x=15\left(5+x\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=4\\x=3\end{matrix}\right.\)

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

pt sau của bạn bị thiếu thì phải

\(\left\{{}\begin{matrix}\left(x+1\right)\left(x^2+1\right)=y^3+1\\\left(y+1\right)\left(y^2+1\right)=z^3+1\\\left(z+1\right)\left(z^2+1\right)=x^3+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^3+x^2+x=y^3\left(1\right)\\y^3+y^2+y=z^3\\z^3+z^2+z=x^3\end{matrix}\right.\)

Giả sử \(x>y\Rightarrow x^3+x^2+x>y^3+y^2+y\)

\(\Rightarrow y^3>z^3\Leftrightarrow y>z\left(2\right)\)

\(\Rightarrow y^3+y^2+y>z^3+z^2+z\Rightarrow z>x\left(3\right)\)

Từ \(\left(2\right);\left(3\right)\Rightarrow y>x\) (Vô lí)

Giả sử \(x< y\Rightarrow x^3+x^2+x< y^3+y^2+y\)

\(\Rightarrow y^3< z^3\Leftrightarrow y< z\left(4\right)\)

\(\Rightarrow y^3+y^2+y< z^3+z^2+z\Rightarrow z< x\left(5\right)\)

Từ \(\left(4\right);\left(5\right)\Rightarrow y< x\) (Vô lí)

\(\Rightarrow x=y=z\)

\(\left(1\right)\Leftrightarrow x^3+x^2+x=x^3\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow x=y=z=0\) hoặc \(x=y=z=-1\)