Rational and Radical Expressions有理表达式与根式表达式

The cancellation law $\dfrac{a c}{b c} = \dfrac{a}{b}$ (for $c \ne 0$) applies to factors joined by multiplication, not terms joined by $+$ or $-$. The classic illegal move is $\dfrac{x + 3}{x} \;\not\to\; 1 + 3 = 4$: the numerator is a sum, not a product. Compare with the legal $\dfrac{x (x + 3)}{x} = x + 3$ (for $x \ne 0$), where $x$ is a true factor. The factor-first rule of §1's first step exists to expose true factors so the cancellation law applies cleanly; forgetting it produces the single most common error in this entire unit.约分律 $\dfrac{a c}{b c} = \dfrac{a}{b}$（$c \ne 0$）只适用于以乘法连接的因式，不适用于以 $+$ 或 $-$ 连接的项。典型的非法操作：$\dfrac{x + 3}{x} \;\not\to\; 1 + 3 = 4$ , 分子是求和，不是乘积。对照合法的 $\dfrac{x (x + 3)}{x} = x + 3$（$x \ne 0$），其中 $x$ 才是真正的因式。§1 第一步的"先分解"规则正是为了暴露真正的因式，让约分律干净地起作用；忘掉它，就会犯下本单元最常见的错误。

For integers, $\operatorname{lcm}(12, 18) = \operatorname{lcm}(2^{2} \cdot 3, \;2 \cdot 3^{2}) = 2^{2} \cdot 3^{2} = 36$ , take the highest power of each prime. The polynomial LCD is the exact same recipe with irreducible polynomials playing the role of primes. If denominators factor as $(x - 1)(x + 2)^{2}$ and $(x - 1)^{2}(x - 3)$, the LCD is $(x - 1)^{2}(x + 2)^{2}(x - 3)$. This is the smallest polynomial that contains both original denominators as factors, which is exactly the property that lets you rewrite each fraction over a single denominator without introducing junk factors.对整数：$\operatorname{lcm}(12, 18) = \operatorname{lcm}(2^{2} \cdot 3, \;2 \cdot 3^{2}) = 2^{2} \cdot 3^{2} = 36$ , 取每个素数的最高次幂。多项式 LCD 配方完全相同，只是不可约多项式扮演"素数"的角色。若分母分别为 $(x - 1)(x + 2)^{2}$ 与 $(x - 1)^{2}(x - 3)$，则 LCD 为 $(x - 1)^{2}(x + 2)^{2}(x - 3)$。这是同时包含两个原分母为因式的最小多项式 , 正是这一性质让我们能把每个分数改写到同一分母上而不引入多余因式。

Multiplying an equation by an expression $E(x)$ produces a new equation whose solution set is the union of (a) solutions of the original and (b) zeros of $E(x)$. If the original has $E(x)$ as a denominator, then zeros of $E(x)$ were excluded; the multiplied equation re-includes them, producing extraneous solutions. Formally: if $A = B$ on a domain $D$, then $A E = B E$ on $D$ trivially , but the polynomial equation $A E = B E$ holds on all of $\mathbb{R}$ wherever $E = 0$ (both sides are $0$), regardless of whether the original was true there. The extra solutions live at $\{x : E(x) = 0\}$, exactly the restriction set. The discard step in §3's algorithm filters them out.把方程两边乘上表达式 $E(x)$ 后，新方程的解集等于原方程的解集与 $E(x)$ 零点集合的并。如果原方程把 $E(x)$ 作为分母，那么 $E(x)$ 的零点本被排除；乘以 $E(x)$ 后这些零点被重新放回，从而产生增根。形式化地：若 $A = B$ 在定义域 $D$ 上成立，则 $A E = B E$ 在 $D$ 上平凡成立 , 但作为多项式方程，$A E = B E$ 在 $E = 0$ 处自动成立（两边都为 $0$），与原方程是否成立无关。多出来的解恰好位于 $\{x : E(x) = 0\}$，即限制集合。§3 算法中的"舍去"步骤正用于过滤掉它们。

If $p(x) = (x - a) \tilde p(x)$ and $q(x) = (x - a) \tilde q(x)$ share the factor $(x - a)$ with $\tilde q(a) \ne 0$, then for $x \ne a$ we cancel: $f(x) = \tilde p(x) / \tilde q(x)$. The simplified expression has a well-defined value $\tilde p(a) / \tilde q(a)$ at $x = a$, but $f$ itself is undefined there because $q(a) = 0$. The graph has a "hole" at $(a, \tilde p(a) / \tilde q(a))$: the surrounding curve approaches the point, but the point itself is missing.若 $p(x) = (x - a) \tilde p(x)$、$q(x) = (x - a) \tilde q(x)$ 共享因式 $(x - a)$ 且 $\tilde q(a) \ne 0$，则在 $x \ne a$ 时可以约分：$f(x) = \tilde p(x) / \tilde q(x)$。化简后的表达式在 $x = a$ 处取良定义的值 $\tilde p(a) / \tilde q(a)$，但 $f$ 本身在那里未定义，因为 $q(a) = 0$。图像在 $(a, \tilde p(a) / \tilde q(a))$ 处有一个"空洞"：周围的曲线趋近该点，但该点本身缺失。

By contrast, if $q(b) = 0$ but $p(b) \ne 0$, no cancellation removes the $(x - b)$ factor. As $x \to b$, the numerator approaches a nonzero constant and the denominator approaches $0$; $|f(x)| \to \infty$. That is a vertical asymptote. This vocabulary is the bridge to IB Math HL B3 (Functions with Asymptotes) and to AP Calc Unit 1, where the same factor-and-cancel move is the standard limits-at-a-point computation for removable discontinuities.对照之下，若 $q(b) = 0$ 而 $p(b) \ne 0$，任何约分都无法消去 $(x - b)$。当 $x \to b$ 时，分子趋于一个非零常数而分母趋于 $0$，故 $|f(x)| \to \infty$ , 这就是垂直渐近线。这套词汇正是通向 IB Math HL B3（含渐近线的函数）与 AP Calc Unit 1 的桥梁，那里同样的"分解—约分"动作是可去间断点处的标准极限计算。

The principal square root is the unique non-negative $y$ with $y^{2} = x$. If $a$ is real, $a^{2}$ is non-negative, but $a$ itself might be negative. So $\sqrt{a^{2}} = |a|$ regardless of the sign of $a$: equals $a$ when $a \ge 0$, equals $-a$ when $a < 0$. This is why writing $\sqrt{(x - 3)^{2}} = x - 3$ is illegal in general , it's only true when $x \ge 3$. The unconditional identity is $\sqrt{(x - 3)^{2}} = |x - 3|$. For odd indices, the issue does not arise: $\sqrt[3]{a^{3}} = a$ for every real $a$ because the cube root preserves sign. The absolute-value subtlety is an even-index-specific phenomenon.主平方根是满足 $y^{2} = x$ 的唯一非负 $y$。若 $a$ 为实数，$a^{2}$ 非负，但 $a$ 本身可能为负。故无论 $a$ 的符号如何，$\sqrt{a^{2}} = |a|$：$a \ge 0$ 时为 $a$；$a < 0$ 时为 $-a$。因此一般而言 $\sqrt{(x - 3)^{2}} = x - 3$ 是非法的 , 只有在 $x \ge 3$ 时成立。无条件成立的恒等式是 $\sqrt{(x - 3)^{2}} = |x - 3|$。奇次根不会出现这个问题：对任意实 $a$，$\sqrt[3]{a^{3}} = a$，因为立方根保号。绝对值的微妙之处仅属于偶次根。

$A = B \Rightarrow A^{2} = B^{2}$ is true; the converse is false: $A^{2} = B^{2}$ only tells us $A = \pm B$. Squaring an equation can introduce solutions of $A = -B$ in addition to those of $A = B$ , the new ones are extraneous because the original didn't include them. For $\sqrt{x + 6} = x$ the principal root is $\ge 0$, so the squared equation $x + 6 = x^{2}$ is equivalent to $\sqrt{x + 6} = x$ or $\sqrt{x + 6} = -x$; the candidate $x = -2$ satisfies the second branch ($\sqrt{4} = 2 = -(-2)$) but not the first. This is the same mechanism as in §3 (multiplying by a denominator): any non-injective operation applied to both sides can re-include excluded values and requires a post-hoc check.$A = B \Rightarrow A^{2} = B^{2}$ 成立；其逆不成立：$A^{2} = B^{2}$ 只能得出 $A = \pm B$。对方程两边平方，可能在 $A = B$ 的解之外引入 $A = -B$ 的解 , 这些新解是多余根，因为原方程并不包含它们。在 $\sqrt{x + 6} = x$ 中主根 $\ge 0$，故平方后的方程 $x + 6 = x^{2}$ 等价于 $\sqrt{x + 6} = x$ 或 $\sqrt{x + 6} = -x$；候选根 $x = -2$ 满足第二支（$\sqrt{4} = 2 = -(-2)$）而不满足第一支。这与 §3（乘以分母）是同一机制：对两边施加任何非单射运算，都可能把被排除的值重新放回，从而需要事后检验。

To extend exponentiation from integer exponents to rational ones while preserving $(a^{p})^{q} = a^{p q}$: $a^{1/n}$ must satisfy $(a^{1/n})^{n} = a^{1} = a$, so $a^{1/n}$ is an $n$-th root of $a$; we pick the principal one, giving $a^{1/n} = \sqrt[n]{a}$. Then $a^{m/n} = (a^{1/n})^{m} = (\sqrt[n]{a})^{m}$, or equivalently $(a^{m})^{1/n} = \sqrt[n]{a^{m}}$. Both forms agree because both are forced by the exponent laws , CCSSM's HSN-RN.A.1 calls this "extending the properties of integer exponents." The same logic forces $a^{-p} = 1/a^{p}$ (from $a^{p} \cdot a^{-p} = a^{0} = 1$) and $a^{0} = 1$ (from $a^{p}/a^{p} = a^{0}$).把指数运算从整数指数推广到有理指数时，若要保留 $(a^{p})^{q} = a^{p q}$：$a^{1/n}$ 必须满足 $(a^{1/n})^{n} = a^{1} = a$，所以 $a^{1/n}$ 是 $a$ 的 $n$ 次方根；取主根得 $a^{1/n} = \sqrt[n]{a}$。再由此推得 $a^{m/n} = (a^{1/n})^{m} = (\sqrt[n]{a})^{m}$，或等价地 $(a^{m})^{1/n} = \sqrt[n]{a^{m}}$。两种形式吻合，因为都是被指数律唯一决定的 , CCSSM HSN-RN.A.1 把这称为"延伸整数指数的运算律"。同一逻辑也强制了 $a^{-p} = 1/a^{p}$（由 $a^{p} \cdot a^{-p} = a^{0} = 1$）与 $a^{0} = 1$（由 $a^{p}/a^{p} = a^{0}$）。