📚 9C Answers | 9C 练习答案与解析
This article provides worked answers and key methods for Exercise 9C, focusing on algebraic manipulation, equations, proportion, Pythagoras' Theorem and basic trigonometry. Use each section to check your final answers and revise the exact steps needed.
本文提供 9C 练习的详细答案与核心方法，涵盖代数运算、方程、比例、勾股定理和基础三角。请逐节核对最终答案，并复习每一步的解题过程。
1. Expanding Single Brackets | 单项式展开括号
Q1 Expand 3x(x + 5). Multiply 3x by x to get 3x² and 3x by 5 to get 15x, so the answer is 3x² + 15x.
Q1 展开 3x(x + 5)。用 3x 乘以 x 得到 3x²，用 3x 乘以 5 得到 15x，因此答案是 3x² + 15x。
expand the bracket: -2 × 4x = -8x and -2 × (-3) = +6. Then add 5x: -8x + 6 + 5x = -3x + 6.
Q2 化简 -2(4x - 3) + 5x。先展开括号：-2 × 4x = -8x，-2 × (-3) = +6。再加上 5x：-8x + 6 + 5x = -3x + 6。
8 - 6x - 3. Collect like terms: 4x - 6x = -2x and -8 - 3 = -11, so the answer is -2x - 11.
Q3 化简 4(x - 2) - 3(2x + 1)。分别展开两个括号：4x - 8 - 6x - 3。合并同类项：4x - 6x = -2x，-8 - 3 = -11，因此答案是 -2x - 11。
Always multiply each term inside the bracket by the factor outside, keeping the sign in front of each product.
务必用括号外的系数乘以括号内的每一项，并保留每一项乘积前面的符号。
2. Expanding Double Brackets | 展开双括号
= -7x, 4 × x = 4x, 4 × (-7) = -28. Combine -7x + 4x = -3x, so the answer is x² - 3x - 28.
Q1 展开 (x + 4)(x - 7)。使用 FOIL 方法：x × x = x²，x × (-7) = -7x，4 × x = 4x，4 × (-7) = -28。合并 -7x + 4x = -3x，因此答案是 x² - 3x - 28。
= -6x, 1 by x = x, and 1 by -3 = -3. Combine -6x + x = -5x, so the result is 2x² - 5x - 3.
Q2 展开 (2x + 1)(x - 3)。先算 2x × x = 2x²，2x × -3 = -6x，1 × x = x，1 × -3 = -3。合并 -6x + x = -5x，因此结果是 2x² - 5x - 3。
3x × 5 = 15x, -2 × 2x = -4x, -2 × 5 = -10. Combine 15x - 4x = 11x, giving 6x² + 11x - 10.
Q3 展开 (3x - 2)(2x + 5)。逐项相乘：3x × 2x = 6x²，3x × 5 = 15x，-2 × 2x = -4x，-2 × 5 = -10。合并 15x - 4x = 11x，得到 6x² + 11x - 10。
After expanding, always collect like terms to write the expression in its simplest form.
展开之后，一定要合并同类项，将表达式写成最简形式。
3. Factorising by Taking Out Common Factors | 提取公因式
Q1 Factorise 6x² + 9x. The highest common factor of 6x² and 9x is 3x. Divide each term by 3x to get 2x and 3, so the answer is 3x(2x + 3).
Q1 因式分解 6x² + 9x。6x² 和 9x 的最大公因式是 3x。将每一项除以 3x，得到 2x 和 3，因此答案是 3x(2x + 3)。
Q2 Factorise 12x³ - 8x² + 4x. The common factor is 4x. Dividing each term gives 3x², -2x and +1, so the expression becomes 4x(3x² - 2x + 1).
Q2 因式分解 12x³ - 8x² + 4x。公因式是 4x。每一项除以 4x 后得到 3x²、-2x 和 +1，因此表达式变为 4x(3x² - 2x + 1)。
Q3 Factorise 5xy + 10x²y. The common factor is 5xy. Divide the first term by 5xy to get 1, and the second term by 5xy to get 2x. The answer is 5xy(1 + 2x).
Q3 因式分解 5xy + 10x²y。公因式是 5xy。第一项除以 5xy 得到 1，第二项除以 5xy 得到 2x。答案是 5xy(1 + 2x)。
Always check your factorisation by expanding the bracket to ensure it matches the original expression.
检查因式分解时，可以重新展开括号，确认结果与原式一致。
4. Factorising Quadratic Trinomials | 二次三项式因式分解
t multiply to 15 and add to 8: these are 3 and 5. Therefore x² + 8x + 15 = (x + 3)(x + 5).
Q1 因式分解 x² + 8x + 15。找出两个数，乘积为 15，和为 8：这两个数是 3 和 5。因此 x² + 8x + 15 = (x + 3)(x + 5)。
Q2 Factorise x² - 5x - 24. Look for two numbers that multiply to -24 and add to -5. The numbers are -8 and 3, so the factors are (x - 8)(x + 3).
Q2 因式分解 x² - 5x - 24。寻找乘积为 -24、和为 -5 的两个数。这两个数是 -8 和 3，因此因式为 (x - 8)(x + 3)。
middle term: 2x² + 6x + x + 3. Factor by grouping: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).
Q3 因式分解 2x² + 7x + 3。将 2 和 3 相乘得到 6。寻找乘积为 6、和为 7 的两个数：它们是 6 和 1。拆分中间项：2x² + 6x + x + 3。分组分解：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。
For quadratics with leading coefficient not equal to 1, use the splitting method or trial and error carefully.
当二次项系数不为 1 时，应使用拆分法或仔细试算。
5. Solving Linear Equations | 解一元一次方程
ct 2x from both sides to get 2x - 7 = 9. Add 7 to both sides: 2x = 16. Divide by 2: x = 8.
Q1 解方程 4x - 7 = 2x + 9。两边同时减去 2x，得到 2x - 7 = 9。两边同时加 7：2x = 16。两边除以 2：x = 8。
ly both sides by 5: 2(x + 3) = 30. Divide both sides by 2: x + 3 = 15. Subtract 3: x = 12.
Q2 解方程 2(x + 3) / 5 = 6。两边同乘 5：2(x + 3) = 30。两边同除以 2：x + 3 = 15。再减去 3：x = 12。
sides: 6x - 3 = 4x + 20. Subtract 4x: 2x - 3 = 20. Add 3: 2x = 23. Divide by 2: x = 11.5.
Q3 解方程 3(2x - 1) = 4(x + 5)。展开两边：6x - 3 = 4x + 20。减去 4x：2x - 3 = 20。加 3：2x = 23。除以 2：x = 11.5。
Always verify your solution by substituting it back into the original equation.
解完方程后，应将解代入原方程进行检验。
6. Solving Quadratic Equations by Factorising | 用因式分解法解二次方程
- 5) = 0. Set each factor equal to zero: x - 2 = 0 gives x = 2, and x - 5 = 0 gives x = 5.
Q1 解方程 x² - 7x + 10 = 0。将左边因式分解：(x - 2)(x - 5) = 0。令每一个因式等于零：x - 2 = 0 得 x = 2，x - 5 = 0 得 x = 5。
as (2x - 1)(x + 3) = 0. Then 2x - 1 = 0 gives x = 1/2 or 0.5, and x + 3 = 0 gives x = -3.
Q2 解方程 2x² + 5x - 3 = 0。因式分解为 (2x - 1)(x + 3) = 0。于是 2x - 1 = 0 得 x = 1/2 或 0.5，x + 3 = 0 得 x = -3。
torise as (x + 6)(x - 2) = 0. Therefore x + 6 = 0 gives x = -6, and x - 2 = 0 gives x = 2.
Q3 解方程 x² + 4x - 12 = 0。因式分解为 (x + 6)(x - 2) = 0。因此 x + 6 = 0 得 x = -6，x - 2 = 0 得 x = 2。
The zero product property is essential: if a product equals zero, at least one factor must be zero.
零乘积性质是关键：如果乘积等于零，那么至少有一个因式等于零。
7. Simplifying Algebraic Fractions | 代数分式化简
Q1 Simplify (x² - 9) / (x + 3). Factorise the numerator as (x - 3)(x + 3). Cancel the common factor (x + 3), noting that x ≠ -3, to get x - 3.
Q1 化简 (x² - 9) / (x + 3)。将分子因式分解为 (x - 3)(x + 3)。约去公因式 (x + 3)，并注意 x ≠ -3，得到 x - 3。
2x and denominator 4x reduce to 2/4 = 1/2. The final answer is 1/2, with x ≠ 0 and x ≠ -1
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