高中数学必修 第一册（648题）

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知识点：第三章 函数的概念与性质

参考答案：\(∀{x}_{1}，{x}_{2}∈(0，1)\)，且\({x}_{1}<{x}_{2}\)，则

\(f\left ( {{x}_{1}} \right )-f\left ( {{x}_{2}} \right )=\left ( {{x}_{1}+\frac {1} {{x}_{1}}} \right )-\left ( {{x}_{2}+\frac {1} {{x}_{2}}} \right )=\left ( {{x}_{1}-{x}_{2}} \right )+\left ( {\frac {1} {{x}_{1}}-\frac {1} {{x}_{2}}} \right )=\left ( {{x}_{1}-{x}_{2}} \right )+\frac {{x}_{2}-{x}_{1}} {{x}_{1}{x}_{2}}=\left ( {{x}_{1}-{x}_{2}} \right )\cdot \frac {{{x}_{1}x}_{2}-1} {{x}_{1}{x}_{2}}\)

\(∵0<{x}_{1}<{x}_{2}<1，∴{x}_{1}-{x}_{2}<0，0<{x}_{1}{x}_{2}<1，{x}_{1}{x}_{2}-1<0\)，\(\therefore f\left ( {{x}_{1}} \right )-f\left ( {{x}_{2}} \right )>0\)即\(f\left ( {{x}_{1}} \right )>f\left ( {{x}_{2}} \right )\)．\(∴f(x)\)在\((0，1)\)上单调递减．