题目内容

4.解方程组:$\left\{\begin{array}{l}{{x}^{2}+{y}^{2}=26}\\{xy=5}\end{array}\right.$.

分析 xy=5,可得x2y2=25,又x2+y2=26,可得x2,y2是一元二次方程t2-26t+25=0的两个实数根,且x,y同号,解出即可.

解答 解:∵xy=5,∴x2y2=25,
又x2+y2=26,
∴x2,y2是一元二次方程t2-26t+25=0的两个实数根.
解得$\left\{\begin{array}{l}{{x}^{2}=1}\\{{y}^{2}=25}\end{array}\right.$或$\left\{\begin{array}{l}{{x}^{2}=25}\\{{y}^{2}=1}\end{array}\right.$,
又x,y同号,
∴$\left\{\begin{array}{l}{x=1}\\{y=5}\end{array}\right.$,$\left\{\begin{array}{l}{x=-1}\\{y=-5}\end{array}\right.$,或$\left\{\begin{array}{l}{x=5}\\{y=1}\end{array}\right.$,$\left\{\begin{array}{l}{x=-5}\\{y=-1}\end{array}\right.$.即为原方程组的解.

点评 本题考查了方程组的解法、一元二次方程的解法,考查了计算能力,属于中档题.

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