Question

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

Answer 1

分析 方程組變形為$\left\{\begin{array}{l}{\frac{1}{a}+\frac{1}=3}\\{\frac{1}+\frac{1}{c}=4}\\{\frac{1}{a}+\frac{1}{c}=5}\end{array}\right.$，設$\frac{1}{a}$=x，$\frac{1}$=y，$\frac{1}{c}$=z，則$\left\{\begin{array}{l}{x+y=3①}\\{y+z=4②}\\{x+z=5③}\end{array}\right.$，利用加減消元法即可求得．

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$\left\{\begin{array}{l}{\frac{1}{a}+\frac{1}=3}\\{\frac{1}+\frac{1}{c}=4}\\{\frac{1}{a}+\frac{1}{c}=5}\end{array}\right.$，
設$\frac{1}{a}$=x，$\frac{1}$=y，$\frac{1}{c}$=z，
則$\left\{\begin{array}{l}{x+y=3①}\\{y+z=4②}\\{x+z=5③}\end{array}\right.$
①-②得x-z=-1④，
與③組成方程組$\left\{\begin{array}{l}{x-z=-1}\\{x+z=5}\end{array}\right.$
解得$\left\{\begin{array}{l}{x=2}\\{z=3}\end{array}\right.$，
把x=2代入①得，y=1，
∴原方程組的解$\left\{\begin{array}{l}{a=\frac{1}{2}}\\{b=1}\\{c=\frac{1}{3}}\end{array}\right.$．

點評 此題考查了解三元一次方程組，利用了換元、消元的思想，消元的方法有：代入消元法與加減消元法．