Question

14．如果$\frac{1}{{s}_{1}}=\frac{1}{{t}_{1}}+\frac{1}{{t}_{2}}$，$\frac{1}{{s}_{2}}=\frac{1}{{t}_{1}}-\frac{1}{{t}_{2}}$，则$\frac{{s}_{1}}{{s}_{2}}$=（　　）

• A． $\frac{{t}_{1}+{t}_{2}}{{t}_{2}-{t}_{1}}$
• B． $\frac{{t}_{2}-{t}_{1}}{{t}_{2}+{t}_{1}}$
• C． $\frac{{t}_{1}-{t}_{2}}{{t}_{2}+{t}_{1}}$
• D． $\frac{{t}_{1}+{t}_{2}}{{t}_{1}-{t}_{2}}$

Answer 1

分析 先化简，求得S₁，S₂，再代入求比值即可．

解答 解：∵$\frac{1}{{s}_{1}}=\frac{1}{{t}_{1}}+\frac{1}{{t}_{2}}$，$\frac{1}{{s}_{2}}=\frac{1}{{t}_{1}}-\frac{1}{{t}_{2}}$，
∴S₁=$\frac{{t}_{1}{t}_{2}}{{t}_{1}+{t}_{2}}$，
S₂=$\frac{{t}_{1}{t}_{2}}{{t}_{2}-{t}_{1}}$，
∴$\frac{{s}_{1}}{{s}_{2}}$=$\frac{\frac{{t}_{1}{t}_{2}}{{t}_{1}+{t}_{2}}}{\frac{{t}_{1}{t}_{2}}{{t}_{2}-{t}_{1}}}$=$\frac{{t}_{2}-{t}_{1}}{{t}_{1}+{t}_{2}}$，
故选B．

点评 本题考查了分式的混合运算，化简S₁，S₂是解题的关键．