Đáp án:

6) (x;y;z) = (0;0;5/4)

Giải thích các bước giải:

\(\eqalign{ & 6)\,\,\left\{ \matrix{ {x^2} + {y^2} = 4z - 5 + 2xy \hfill \cr {x^4} + {y^4} = 9z - 5 - 4{z^2} - 2{x^2}{y^2} \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ {x^2} + {y^2} - 2xy = 4z - 5 \hfill \cr {x^4} + {y^4} + 2{x^2}{y^2} = 9z - 5 - 4{z^2} \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ {\left( {x - y} \right)^2} = 4z - 5\,\,\left( 1 \right) \hfill \cr {\left( {{x^2} + {y^2}} \right)^2} = 9z - 5 - 4{z^2}\,\,\left( 2 \right) \hfill \cr} \right.\,\,\left( * \right) \cr & \left( 1 \right) \Leftrightarrow 4z - 5 \ge 0 \Leftrightarrow z \ge {5 \over 4} \cr & \left( 2 \right) \Leftrightarrow 9z - 5 - 4{z^2} \ge 0 \Leftrightarrow 1 \le z \le {5 \over 4} \cr & \Rightarrow z = {5 \over 4} \cr & \left( * \right) \Rightarrow \left\{ \matrix{ {\left( {x - y} \right)^2} = 0 \hfill \cr {\left( {{x^2} + {y^2}} \right)^2} = 0 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ x - y = 0 \hfill \cr x = y = 0 \hfill \cr} \right. \Leftrightarrow x = y = 0 \cr} \) Vậy hpt có nghiệm (x;y;z) = (0;0;5/4)