Question

10．已知方程組$\left\{\begin{array}{l}{{x}^{2}+2{y}^{2}=m}\\{3xy=n}\end{array}\right.$的一組解為$\left\{\begin{array}{l}{{x}_{1}=3}\\{{y}_{1}=-2}\end{array}\right.$，則這個(gè)方程組的其他解為$\left\{\begin{array}{l}{{x}_{1}=-3}\\{{y}_{1}=2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=2\sqrt{2}}\\{{y}_{2}=-\frac{3}{2}\sqrt{2}}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{3}=-2\sqrt{2}}\\{{y}_{3}=\frac{3}{2}\sqrt{2}}\end{array}\right.$．

Answer 1

分析 把方程組的一組解代入方程組求出m、n的值，再解原方程組，即可求出這個(gè)方程組的其他解．

解答 解：把$\left\{\begin{array}{l}{{x}_{1}=3}\\{{y}_{1}=-2}\end{array}\right.$代入方程組得，m=17，n=-18，
原方程組為：$\left\{\begin{array}{l}{{x}^{2}+2{y}^{2}=17①}\\{3xy=-18②}\end{array}\right.$，
由②得，x=-$\frac{6}{y}$③，
把③代入①得，2y⁴-17y²+36=0，
解得：y₁=2，y₂=-2，y₃=$\frac{3}{2}\sqrt{2}$，y₄=-$\frac{3}{2}\sqrt{2}$，
把：y₁=2，y₂=-2，y₃=$\frac{3}{2}\sqrt{2}$，y₄=-$\frac{3}{2}\sqrt{2}$代入③得這個(gè)方程組的其他解：
$\left\{\begin{array}{l}{{x}_{1}=-3}\\{{y}_{1}=2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=2\sqrt{2}}\\{{y}_{2}=-\frac{3}{2}\sqrt{2}}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{3}=-2\sqrt{2}}\\{{y}_{3}=\frac{3}{2}\sqrt{2}}\end{array}\right.$．
故答案為：$\left\{\begin{array}{l}{{x}_{1}=-3}\\{{y}_{1}=2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=2\sqrt{2}}\\{{y}_{2}=-\frac{3}{2}\sqrt{2}}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{3}=-2\sqrt{2}}\\{{y}_{3}=\frac{3}{2}\sqrt{2}}\end{array}\right.$．

點(diǎn)評(píng) 本題考查的是高次方程的解法，把已知的一組解代入原方程組求出m、n的值是解題的關(guān)鍵，解答時(shí)，注意代入法的正確運(yùn)用．