题目内容
5.解方程组:$\left\{\begin{array}{l}{{x}^{3}+{x}^{3}{y}^{3}+{y}^{3}=12}\\{x+xy+y=0}\end{array}\right.$.分析 把已知方程组第一个方程变形,结合第二个方程求出xy的值,进一步求出x+y的值,再联立方程组求解.
解答 解:由x3+x3y3+y3=12,得x3+x3y3+y3+1=13,即(x3+1)(y3+1)=13
∴(x+1)(x2-x+1)(y+1)(y2-y+1)=13.
由x+y+xy=0,得x+y+xy+1=1,即(x+1)(y+1)=1,
∴有(x2-x+1)(y2-y+1)=13,
则[(x+1)2-3x][(y+1)2-3y]=13,
(x+1)2(y+1)2-3x(y+1)2-3y(x+1)2+9xy=13,
即1-3x(y2+2y+1)-3y(x2+2x+1)+9xy=13,
∴1-3xy(x+y)-3(x+y)-3xy=1-3xy(1-xy)-3(1-xy)-3xy=13,
x2y2-xy-5=0,
求得:xy=$\frac{1±\sqrt{21}}{2}$,带入x+xy+y=0,可求得:x+y=$\frac{-1-\sqrt{21}}{2}$或x+y=$\frac{-1+\sqrt{21}}{2}$.
联立$\left\{\begin{array}{l}{xy=\frac{1+\sqrt{21}}{2}}\\{x+y=\frac{-1-\sqrt{21}}{2}}\end{array}\right.$,此方程无解;
联立$\left\{\begin{array}{l}{xy=\frac{1-\sqrt{21}}{2}}\\{x+y=\frac{-1+\sqrt{21}}{2}}\end{array}\right.$,解得$\left\{\begin{array}{l}{x=\frac{\sqrt{21}+\sqrt{14+2\sqrt{21}}+1}{4}}\\{y=\frac{-\sqrt{21}+\sqrt{14+2\sqrt{21}}+3}{4}}\end{array}\right.$或$\left\{\begin{array}{l}{x=\frac{\sqrt{21}-\sqrt{14+\sqrt{21}}-1}{4}}\\{y=\frac{-\sqrt{21}-\sqrt{14+2\sqrt{21}}+1}{4}}\end{array}\right.$.
点评 本题考查有理指数幂的化简与求值,考查数学转化思想方法,考查了灵活变形能力和运算能力,难度较大.
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