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题目
题目：
设
\[ f\left (u+\frac {1}{u}\right )=\frac {u+u^3}{1+u^4}, \]
则当 \(x\ge 2\) 时，求 \(f(x)\)。
题面校注： 用户图片中分子首项疑似为 \(-u\)，但图片所给参考答案 \(\frac {x}{x^2-2}\) 对应的是分子为 \(u+u^3\) 的版本。若分子确认为 \(-u+u^3\)，则还需要指定 \(u\) 的取值分支，否则 \(f\left
(u+\frac {1}{u}\right )\) 不能唯一确定。
题解
思路入口：
目标是求 \(f(x)\)，已知的是 \(f\left (u+\frac {1}{u}\right )\)。所以不要把左右两边的 \(x\) 混用，应先把原式中的自变量记作 \(u\)，再把
\[ t=u+\frac {1}{u} \]
看成一个整体。
详细解答：
由题设
\[ f\left (u+\frac {1}{u}\right )=\frac {u+u^3}{1+u^4}. \]
为了把右边也化成 \(u+\frac {1}{u}\) 的形式，对分子、分母同时除以 \(u^2\)。这里要求 \(u\ne 0\)，而 \(u+\frac {1}{u}\) 本身也只在 \(u\ne 0\) 时有意义：
\[ \frac {u+u^3}{1+u^4} =\frac {\frac {1}{u}+u}{\frac {1}{u^2}+u^2}. \]
又因为
\[ u^2+\frac {1}{u^2} =\left (u+\frac {1}{u}\right )^2-2, \]
所以
\[ f\left (u+\frac {1}{u}\right ) =\frac {u+\frac {1}{u}}{\left (u+\frac {1}{u}\right )^2-2}. \]
令
\[ t=u+\frac {1}{u}, \]
则
\[ f(t)=\frac {t}{t^2-2}. \]
题目要求 \(x\ge 2\) 时的 \(f(x)\)，于是把 \(t\) 换回 \(x\)，得到
\[ f(x)=\frac {x}{x^2-2}\qquad (x\ge 2). \]
补充解法：反解参数后消元
也可以先把最终自变量记为 \(t\)，令
\[ t=u+\frac {1}{u}\qquad (t\ge 2). \]
这等价于
\[ u^2-tu+1=0. \]
由此可得
\[ 1+u^2=tu. \]
于是
\[ u+u^3=u(1+u^2)=tu^2. \]
另一方面，由
\[ u^2+\frac {1}{u^2}=\left (u+\frac {1}{u}\right )^2-2=t^2-2 \]
可得
\[ 1+u^4=u^2(t^2-2). \]
所以
\[ f(t)=\frac {u+u^3}{1+u^4} =\frac {tu^2}{u^2(t^2-2)} =\frac {t}{t^2-2}. \]
因此
\[ f(x)=\frac {x}{x^2-2}\qquad (x\ge 2). \]
关于显式解出 \(u\) 再代回：
也可以直接解二次方程
\[ u^2-tu+1=0, \]
得到
\[ u=\frac {t\pm \sqrt {t^2-4}}{2}. \]
但这里不能随意写成单值的 \(u=\theta (t)\)，因为当 \(t>2\) 时有两个分支，它们互为倒数。若将 \(u\) 换成 \(\frac {1}{u}\)，则
\[ \frac {\frac {1}{u}+\frac {1}{u^3}}{1+\frac {1}{u^4}} =\frac {u^3+u}{u^4+1} =\frac {u+u^3}{1+u^4}, \]
所以本题两个分支代回右端得到同一个值，显式解根代回是可行的。只是直接代入根式会使运算变繁，考场上不如先用方程关系消元。
方法比较： 本法本质上是先由 \(t=u+\frac {1}{u}\) 建立 \(u\) 与 \(t\) 的方程关系，再用方程消去 \(u\)。它能做，但比整体换元多了一层消元；考场上优先使用前一种整体改写法。若要显式写 \(u=\theta (t)\)，
必须先说明分支对右端函数值没有影响。
答案：
\[ \boxed {f(x)=\frac {x}{x^2-2}\quad (x\ge 2)}. \]
知识点
核心知识点：
已知 \(f(\varphi (u))\) 时，关键不是先解出 \(u\)，而是把右端改写成只含 \(\varphi (u)\) 的形式。本题中
\[ u^2+\frac {1}{u^2}=\left (u+\frac {1}{u}\right )^2-2 \]
是主要变形入口。
可迁移方法：
遇到 \(u+\frac {1}{u}\)、\(u-\frac {1}{u}\)、\(u^2+\frac {1}{u^2}\) 同时出现的题，优先考虑整体换元。若题目限定 \(x\ge 2\)，通常是在提示 \(x\) 可以作为 \(u+\frac {1}{u}\) 的函数值。
易错点
易错点：
-
• 把题设中的参数 \(u\) 和最终要求的自变量 \(x\) 混成同一个变量。
-
• 只想到解方程 \(x=u+\frac {1}{u}\)，却忽略右端可以直接化成整体变量。
-
• 忘记检查题面符号：若分子是 \(-u+u^3\)，则右端为 \(\frac {u-\frac {1}{u}}{u^2+\frac {1}{u^2}}\)，不能直接得到参考答案。
复盘提醒：
看到 \(f\left (u+\frac {1}{u}\right )\) 这类题，先重命名参数，再问自己：右端能不能通过除以某个 \(u^k\) 造出 \(u+\frac {1}{u}\)？