Question

20．解方程組：$\left\{\begin{array}{l}{\sqrt{x+\frac{1}{y}}+\sqrt{x-y-2}=4}\\{2x-y+\frac{1}{y}=10}\end{array}\right.$．

Answer 1

分析 設(shè)$\sqrt{x+\frac{1}{y}}$=a，$\sqrt{x-y-2}$=b，則原方程組化為$\left\{\begin{array}{l}{a+b=4①}\\{{a}^{2}+^{2}=8②}\end{array}\right.$，求出方程組的解，再代入求出x、y即可．

解答 解：設(shè)$\sqrt{x+\frac{1}{y}}$=a，$\sqrt{x-y-2}$=b，
則原方程組化為：$\left\{\begin{array}{l}{a+b=4①}\\{{a}^{2}+^{2}=8②}\end{array}\right.$
由①得：a=4-b③，
把③代入②得：（4-b）²+b²=8，
解得：b₁=b₂=2，
當(dāng)b=2時(shí)，a=2，
即$\left\{\begin{array}{l}{\sqrt{x+\frac{1}{y}}=2}\\{\sqrt{x-y-2}=2}\end{array}\right.$
解得：$\left\{\begin{array}{l}{x=5}\\{y=-1}\end{array}\right.$，
經(jīng)檢驗(yàn)$\left\{\begin{array}{l}{x=5}\\{y=-1}\end{array}\right.$是原方程組的解，
所以原方程組的解為：$\left\{\begin{array}{l}{x=5}\\{y=-1}\end{array}\right.$．

點(diǎn)評(píng) 本題考查了解無理方程組的應(yīng)用，能把無理方程組轉(zhuǎn)化成有理方程組是解此題的關(guān)鍵．