Question

2．方程組$\left\{\begin{array}{l}{\sqrt{2}x+\sqrt{3}y=1}\\{\sqrt{3}x+\sqrt{2}y=2}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=2\sqrt{3}-\sqrt{2}}\\{y=\sqrt{3}-2\sqrt{2}}\end{array}\right.$．

Answer 1

分析 利用解方程組的方法與步驟求得方程組的解即可．

解答 解：$\left\{\begin{array}{l}{\sqrt{2}x+\sqrt{3}y=1①}\\{\sqrt{3}x+\sqrt{2}y=2②}\end{array}\right.$
①×$\sqrt{3}$-②×$\sqrt{2}$得，y=$\sqrt{3}$-2$\sqrt{2}$③，
把③代入①得，$\sqrt{2}$x+$\sqrt{3}$（$\sqrt{3}$-2$\sqrt{2}$）=1，
解得x=2$\sqrt{3}$-$\sqrt{2}$，
所以原方程組的解為$\left\{\begin{array}{l}{x=2\sqrt{3}-\sqrt{2}}\\{y=\sqrt{3}-2\sqrt{2}}\end{array}\right.$．
故答案為：$\left\{\begin{array}{l}{x=2\sqrt{3}-\sqrt{2}}\\{y=\sqrt{3}-2\sqrt{2}}\end{array}\right.$．

點(diǎn)評(píng) 此題考查二次根式的實(shí)際運(yùn)用，掌握二元一次方程組的解法是解決問題的關(guān)鍵．