Question

14．解方程组：$\left\{\begin{array}{l}{\frac{ab}{a+b}=\frac{1}{3}}\\{\frac{bc}{b+c}=\frac{1}{4}}\\{\frac{ca}{c+a}=\frac{1}{5}}\end{array}\right.$．

Answer 1

分析 方程组变形为$\left\{\begin{array}{l}{\frac{1}{a}+\frac{1}{b}=3}\\{\frac{1}{b}+\frac{1}{c}=4}\\{\frac{1}{a}+\frac{1}{c}=5}\end{array}\right.$，设$\frac{1}{a}$=x，$\frac{1}{b}$=y，$\frac{1}{c}$=z，则$\left\{\begin{array}{l}{x+y=3①}\\{y+z=4②}\\{x+z=5③}\end{array}\right.$，利用加减消元法即可求得．

解答 解：由：$\left\{\begin{array}{l}{\frac{ab}{a+b}=\frac{1}{3}}\\{\frac{bc}{b+c}=\frac{1}{4}}\\{\frac{ca}{c+a}=\frac{1}{5}}\end{array}\right.$可知$\left\{\begin{array}{l}{\frac{1}{a}+\frac{1}{b}=3}\\{\frac{1}{b}+\frac{1}{c}=4}\\{\frac{1}{a}+\frac{1}{c}=5}\end{array}\right.$，
设$\frac{1}{a}$=x，$\frac{1}{b}$=y，$\frac{1}{c}$=z，
则$\left\{\begin{array}{l}{x+y=3①}\\{y+z=4②}\\{x+z=5③}\end{array}\right.$
①-②得x-z=-1④，
与③组成方程组$\left\{\begin{array}{l}{x-z=-1}\\{x+z=5}\end{array}\right.$
解得$\left\{\begin{array}{l}{x=2}\\{z=3}\end{array}\right.$，
把x=2代入①得，y=1，
∴原方程组的解$\left\{\begin{array}{l}{a=\frac{1}{2}}\\{b=1}\\{c=\frac{1}{3}}\end{array}\right.$．

点评 此题考查了解三元一次方程组，利用了换元、消元的思想，消元的方法有：代入消元法与加减消元法．