\(\left\{{}\begin{matrix}x\left(x+y+z\right)=7\\y\left(x+y+z\right)=3\\z\left(x+y+z\right)=15\end{matrix}\right.\) (đoán là đề vậy thôi vì bạn viết thiếu)
\(\Rightarrow x\left(x+y+z\right)+y\left(x+y+z\right)+z\left(x+y+z\right)=7+3+15\)\(\Rightarrow\left(x+y+z\right)\left(x+y+z\right)=25\)
\(\Rightarrow\left(x+y+z\right)^2=25\)
\(\Rightarrow x+y+z=\pm5\)
....
\(\left\{{}\begin{matrix}xy=\dfrac{6}{7}\\yz=\dfrac{7}{12}\\xz=2\end{matrix}\right.\)
\(\Rightarrow xy.yz.xz=\dfrac{6}{7}.\dfrac{7}{12}.2\)
\(\Rightarrow xyz^2=1\)
\(\Rightarrow xyz=\pm1\)
...
tìm x,y,z
a,x(x+y+z)=7,y(x+y+z)=3,z(x+y+z)x(x+y+z)=7,y(x+y+z)=3,z(x+y+z)=6
b,xy=67,yz=712,xz=2
