📚 Solving Rational Equations and Applications | 分式方程的解法与应用
A rational equation is an equation that contains at least one fraction whose numerator or denominator contains a variable. These equations appear frequently in algebra and in real-world problems involving rates, proportions, and ratios. Mastering them requires careful attention to domain restrictions and a reliable method for clearing denominators.
分式方程是含有至少一个分子或分母中包含变量的分式方程。这类方程在代数以及涉及速率、比例和比率的实际问题中经常出现。掌握分式方程需要特别注意定义域的限制，并熟练运用去分母的方法。
1. Introduction to Rational Equations | 分式方程简介
A rational equation has the form P(x)Q(x) = R(x) t one denominator. For example, 2/(x+1) = 3 is a rational equation, but 2x + 1 = 3 is not.
分式方程具有 P(x)/Q(x) = R(x) 的形式，其中变量至少出现在一个分母中。例如，2/(x+1) = 3 是分式方程，而 2x + 1 = 3 不是。
2. Domain and Undefined Values | 定义域与无定义值
Since division by zero is undefined, any value of the variable that makes a denominator equal to zero cannot be part of the solution set. These forbidden values are excluded from the domain of the equation.
因为除以零是没有定义的，任何使分母等于零的变量值都不能是解集的一部分。这些被禁止的值被排除在方程的定义域之外。
(x-1) = 3, the value x = 1 makes the denominator zero, so we require x ≠ 1 before solving.
例如，在方程 (x+2)/(x-1) = 3 中，x = 1 会使分母为零，所以求解前必须要求 x ≠ 1。
3. Basic Strategy: Clearing Denominators | 基本策略：去分母
The standard method is to multiply both sides of the equation by the least common multiple (LCM) of all denominators. This eliminates the fractions and produces a polynomial equation that can be solved using familiar techniques.
标准方法是在方程两边同时乘以所有分母的最小公倍数（LCM）。这样可以消去分式，得到可以用熟悉的方法求解的整式方程。
2/(x+1) = 3
Multiply both sides by (x+1) to obtain 2 = 3(x+1). Then 2 = 3x + 3, so x = -1/3.
两边同时乘以 (x+1)，得到 2 = 3(x+1)。于是 2 = 3x + 3，所以 x = -1/3。
4. Cross-Multiplication for Proportions | 比例交叉相乘
hortcut for clearing denominators, but it may only be used when both sides of the equation are single fractions.
当方程可以写成简单比例 a/b = c/d 时，交叉相乘得到 ad = bc。这是去分母的一种快捷方式，但只能在方程两边都是单个分式时使用。
(x-1)/(x+2) = 3/4
ing and solving gives 4x - 4 = 3x + 6, hence x = 10. This value is allowed because x ≠ -2.
交叉相乘得到 4(x-1) = 3(x+2)。展开并求解：4x - 4 = 3x + 6，因此 x = 10。该值在定义域内，因为 x ≠ -2。
5. Extraneous Roots | 增根
When we multiply both sides by an expression containing a variable, we may introduce extraneous roots. These are values that satisfy the transformed equation but not the original equation, often because they make a denominator zero.
当我们在方程两边同时乘以含有变量的表达式时，可能会引入增根。增根是满足变换后方程但不满足原方程的值，通常因为它们会使分母为零。
(x²-4)/(x-2) = 1
original denominator zero, so it is an extraneous root. The only valid solution is x = -1.
两边同时乘以 (x-2)，得到 x² - 4 = x - 2。整理得 x² - x - 2 = 0，即 (x-2)(x+1) = 0。得到的候选解为 x = 2 和 x = -1。但 x = 2 会使原方程分母为零，因此是增根。唯一有效的解是 x = -1。
6. Solving Equations with Multiple Denominators | 含多个分母的方程
When an equation contains several rational terms, we find the LCM of all denominators and multiply the entire equation by this LCM. This removes every denominator at once.
当方程包含多个分式项时，我们找出所有分母的最小公倍数，并将方程整体乘以这个 LCM。这样可一次性去除所有分母。
2/x + 3/(x+1) = 1
1). This simplifies to 2x + 2 + 3x = x² + x, so x² - 4x - 2 = 0. The solutions are x = 2 ± √6. Both values are valid because neither equals 0 or -1.
最小公倍数为 x(x+1)。两边同乘得到 2(x+1) + 3x = x(x+1)。化简为 2x + 2 + 3x = x² + x，即 x² - 4x - 2 = 0。解为 x = 2 ± √6。两个值都不等于 0 或 -1，因此都在定义域内。
7. Applications: Work Problems | 应用：工程问题
Work problems often model the rate of completing a task by the fraction of the task done per unit time. If a person or machine can complete a job in a hours, their rate is 1/a per hour.
工程问题常用单位时间内完成工作量的分数来表示工作效率。如果一个人或一台机器能在 a 小时内完成一项工作，那么其效率是每小时完成 1/a。
1/4 + 1/6 = 1/t
hours needed. Then 1/4 + 1/6 = 1/t. Since 1/4 + 1/6 = 5/12, we have t = 12/5 = 2.4 hours.
例如：一根水管单独注满水池需要 4 小时，另一根单独注满需要 6 小时。如果两根水管同时工作，设所需时间为 t 小时。则 1/4 + 1/6 = 1/t。因为 1/4 + 1/6 = 5/12，所以 t = 12/5 = 2.4 小时。
8. Applications: Distance, Rate, and Time | 应用：行程问题
speed relationship. The basic formula is distance = rate × time, so time = distance/rate.
行程问题经常在两次运动的时间或速度关系相同时产生分式方程。基本公式是：距离 = 速度 × 时间，因此时间 = 距离/速度。
45/(x+3) = 30/(x-3)
+3), so 45x - 135 = 30x + 90, hence x = 15. The domain requires x > 3, so x = 15 is valid.
例如：一艘船顺流行驶 45 km 和逆流行驶 30 km 所用时间相同。水流速度为 3 km/h，船在静水中的速度为 x km/h。则顺流速度为 x+3，逆流速度为 x-3。通过交叉相乘解方程得 45(x-3) = 30(x+3)，即 45x - 135 = 30x + 90，因此 x = 15。定义域要求 x > 3，所以 x = 15 是有效解。
9. Applications: Mixture and Ratio Problems | 应用：混合与比例问题
Rational equations also arise when mixing two quantities or adjusting a ratio. In such problems, it is helpful to express the relevant concentration or ratio as a fraction.
分式方程也常出现在混合两种量或调整比例的问题中。解决这类问题时，将相关浓度或比例表示成分式会很有帮助。
5/(10+x) = 0.20
Example: How many litres of pure water must be added to 10 litres of a 50% salt solution to obtain a 20% salt solution? so 5/(10+x) = 0.20. Multiplying through gives 5 = 0.20(10+x) = 2 + 0.2x, so x = 15 litres.
例如：需要向 10 升 50% 的盐溶液中加入多少升纯水，才能得到 20% 的盐溶液？盐的质量仍为 5 升。若加入 x 升水，新的总体积为 10+x，因此 5/(10+x) = 0.20。两边整理得 5 = 0.20(10+x) = 2 + 0.2x，所以 x = 15 升。
10. Common Mistakes to Avoid | 常见错误避免
Forgetting to state the domain before solving.
忘记在求解前确定定义域。
Forgetting to substitute candidates back into the original equation to reject extraneous roots.
忘记将候选解代入原方程检验，从而未能排除增根。
Cross-multiplying when the equation is not a single proportion on both sides.
当方程两边不是单个分式时盲目使用交叉相乘。
Making sign errors when distributing a negative denominator after multiplication.
乘以分母后展开时出现符号错误。
Dividing both sides by a variable expression that could be zero.
两边同时除以可能为零的变量表达式。
11. Practice Questions | 练习题
Solve (2x-1)/(x+3) = 1.
解方程 (2x-1)/(x+3) = 1。
Solve 3/(x-1) = 5/(x+2).
解方程 3/(x-1) = 5/(x+2)。
Solve x/(x-3) - 2 = 3/(x-3).
解方程 x/(x-3) - 2 = 3/(x-3)。
Solve 1/x + 1/(x+2) = 5/(x²+2x).
解方程 1/x + 1/(x+2) = 5/(x²+2x)。
12. Summary | 总结
To solve a rational equation, follow these steps: first determine the domain by excluding values that make any denominator zero; second, clear denominators by multiplying both sides by the LCM; third, solve the resulting polynomial equation; finally, check every candidate solution in the original equation and discard any extraneous roots. With this method, rational equations can be solved accurately and applied to a wide range of practical problems.
解分式方程的基本步骤是：先确定定义域，排除使分母为零的值；再两边同乘最小公倍数，去分母；然后解所得的整式方程；最后将候选解代入原方程检验，舍去增根。掌握这一方法，就能准确求解分式方程，并将其应用于各类实际问题。
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