Question

4．解方程組$\left\{\begin{array}{l}{{x}^{2}+{y}^{2}=52}\\{xy+x+y=34}\end{array}\right.$．

Answer 1

分析 令x+y=a，xy=b，可得x²+y²=（x+y）²-2xy=a²-2b，原方程組變?yōu)椋?\left\{\begin{array}{l}{{a}^{2}-2b=52}\\{a+b=34}\end{array}\right.$．由a+b=34可得：b=34-a，代入a²-2b=52，解得a，b，再利用一元二次方程的根與系數(shù)的關(guān)系即可得出．

解答 解：令x+y=a，xy=b，
則x²+y²=（x+y）²-2xy=a²-2b，
∴原方程組變?yōu)椋?\left\{\begin{array}{l}{{a}^{2}-2b=52}\\{a+b=34}\end{array}\right.$，
由a+b=34可得：b=34-a，
代入a²-2b=52，化為a²+2a-120=0，
解得a=10，或a=-12．
∴$\left\{\begin{array}{l}{a=10}\\{b=24}\end{array}\right.$，或$\left\{\begin{array}{l}{a=-12}\\{b=46}\end{array}\right.$．
∴$\left\{\begin{array}{l}{x+y=10}\\{xy=24}\end{array}\right.$，或$\left\{\begin{array}{l}{x+y=-12}\\{xy=46}\end{array}\right.$，
由此可把x，y分別看做以下一元二次方程的兩個(gè)實(shí)數(shù)根：
t²-10t+24=0，t²+12t-46=0，
分別解得t=4，6；t=-6±$\sqrt{82}$．
∴原方程組的解為$\left\{\begin{array}{l}{x=4}\\{y=6}\end{array}\right.$，$\left\{\begin{array}{l}{x=6}\\{y=4}\end{array}\right.$，$\left\{\begin{array}{l}{x=-6+\sqrt{82}}\\{y=-6-\sqrt{82}}\end{array}\right.$，$\left\{\begin{array}{l}{x=-6-\sqrt{82}}\\{y=-6+\sqrt{82}}\end{array}\right.$．

點(diǎn)評(píng) 本題考查了一元二次方程的根與系數(shù)的關(guān)系、方程組的解法、換元法，考查了推理能力與計(jì)算能力，屬于中檔題．