题目内容
1.解方程组:$\left\{\begin{array}{l}\frac{4}{x+y}+\frac{6}{x-y}=3\\ \frac{9}{x-y}-\frac{1}{x+y}=1\end{array}\right.$.分析 方程组换元整理后,利用加减消元法求出解即可.
解答 解:设$\frac{1}{x+y}$=u,$\frac{1}{x-y}$=v,
方程组整理得:$\left\{\begin{array}{l}{4u+6v=3①}\\{9v-u=1②}\end{array}\right.$,
①+②×4得:42v=7,即v=$\frac{1}{6}$,
把v=$\frac{1}{6}$代入①得:u=$\frac{1}{2}$,
∴$\left\{\begin{array}{l}{\frac{1}{x+y}=\frac{1}{2}}\\{\frac{1}{x-y}=\frac{1}{6}}\end{array}\right.$,即$\left\{\begin{array}{l}{x+y=2}\\{x-y=6}\end{array}\right.$,
解得:$\left\{\begin{array}{l}{x=4}\\{y=-2}\end{array}\right.$,
经检验$\left\{\begin{array}{l}{x=4}\\{y=-2}\end{array}\right.$是方程组的解.
点评 此题考查了解二元一次方程组,利用了换元的思想,熟练掌握运算法则是解本题的关键.
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