Question

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

Answer 1

分析 設(shè)$\frac{1}{x-y}$=a，$\frac{1}{x+y}$=b，方程組變形后求出解得到a與b的值，即可確定出x與y的值．

解答 解：設(shè)$\frac{1}{x-y}$=a，$\frac{1}{x+y}$=b，
方程組變形得：$\left\{\begin{array}{l}{2a+b=\frac{1}{4}①}\\{b-a=1②}\end{array}\right.$，
①-②得：3a=-$\frac{3}{4}$，即a=-$\frac{1}{4}$，
把a(bǔ)=-$\frac{1}{4}$代入②得：b=$\frac{3}{4}$，
可得$\left\{\begin{array}{l}{x-y=-4}\\{x+y=\frac{4}{3}}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{x=-\frac{4}{3}}\\{y=\frac{8}{3}}\end{array}\right.$，
則經(jīng)檢驗(yàn)$\left\{\begin{array}{l}{x=-\frac{4}{3}}\\{y=\frac{8}{3}}\end{array}\right.$是方程組的解．

點(diǎn)評(píng) 此題考查了解二元一次方程組，利用了消元的思想，消元的方法有：代入消元法與加減消元法．