Question

12．解方程组：$\left\{\begin{array}{l}{x+y+z=7}\\{{x}^{2}+{y}^{2}+{z}^{2}=21}\\{xy=8}\end{array}\right.$．

Answer 1

分析 把z=7-（x+y），及xy=8代入x²+y²+z²=21，化为x+y=1或x+y=6．联立①$\left\{\begin{array}{l}{x+y=1}\\{xy=8}\end{array}\right.$，或②$\left\{\begin{array}{l}{x+y=6}\\{xy=8}\end{array}\right.$，分别解出即可．

解答 解：把z=7-（x+y），及xy=8代入x²+y²+z²=21，化为x+y=1或x+y=6．
联立①$\left\{\begin{array}{l}{x+y=1}\\{xy=8}\end{array}\right.$，或②$\left\{\begin{array}{l}{x+y=6}\\{xy=8}\end{array}\right.$，
①无解，②解得$\left\{\begin{array}{l}{x=2}\\{y=4}\end{array}\right.$或$\left\{\begin{array}{l}{x=4}\\{y=2}\end{array}\right.$．
∴$\left\{\begin{array}{l}{x=2}\\{y=4}\\{x=1}\end{array}\right.$或$\left\{\begin{array}{l}{x=4}\\{y=2}\\{z=1}\end{array}\right.$．
∴原方程组的解为：$\left\{\begin{array}{l}{x=2}\\{y=4}\\{x=1}\end{array}\right.$或$\left\{\begin{array}{l}{x=4}\\{y=2}\\{z=1}\end{array}\right.$．

点评 本题考查了方程组的解法，考查了推理能力与计算能力，属于中档题．