Giải các hệ phương trình sau:
$\lceil\,\,1\,\,\rfloor$
$$\left\{\begin{matrix} y+ z+ xz^{\,2}+ 2\,xyz & = & 2\\ 3\,xyz+ 8\,y & = & 1\\ 4\,y+ 2\,z+ 2\,xyz+ 2\,xz^{\,2} & = & 3 \end{matrix}\right.$$
$\lceil\,\,2\,\,\rfloor$
$$\left\{\begin{matrix} 7\,y+ 7\,z+ 7\,xy^{\,2}+ 7\,xyz & = & 6\\ 7\,xyz+ 3\,y+ 3\,z & = & 7\\ 9\,y+ 9\,z+ 6\,xy^{\,2} & = & 8 \end{matrix}\right.$$
$\lceil\,\,3\,\,\rfloor$
$$\left\{\begin{matrix} 2\,xy+ 6\,x+ 6\,z^{\,2}+ 6\,zx & = & 1\\ 7\,xyz+ xy & = & 8\\ 3\,x+ 3\,z^{\,2}+ 3\,zx+ xyz & = & 3 \end{matrix}\right.$$
$\lceil\,\,4\,\,\rfloor$
$$\left\{\begin{matrix} 6\,z+ 6\,x+ 6\,zx+ 6\,x^{\,2}+ 4\,xyz & = & 5\\ 12\,x+ 7\,z+ 7\,zx+ 7\,x^{\,2} & = & 2\\ 3\,xyz+ 2\,x & = & 3 \end{matrix}\right.$$
$\lceil\,\,5\,\,\rfloor$
$$\left\{\begin{matrix} 4\,zx+ 7\,xy+ 7\,zx+ 7\,y^{\,2} & = & 3\\ 4\,xyz+ zx & = & 2\\ 2\,xy+ 2\,zx+ 2\,y^{\,2}+ 5\,xyz & = & 4 \end{matrix}\right.$$
$\lceil\,\,6\,\,\rfloor$
$$\left\{\begin{matrix} 3\,zx+ 6\,xy^{\,2}+ 6\,xyz & = & 1\\ 3\,xy^{\,2}+ 4\,xyz & = & 3\\ 7\,xyz+ 6\,zx & = & 8 \end{matrix}\right.$$
$\lceil\,\,7\,\,\rfloor$
$$\left\{\begin{matrix} 8\,xz+ 8\,zx^{\,2}+ 8\,z^{\,2}x^{\,2} & = & 1\\ 5\,xyz+ 3\,zx & = & 2\\ 8\,zx^{\,2}+ 8\,x^{\,2}z^{\,2}+ 8\,xyz & = & 5 \end{matrix}\right.$$