📚 Answers pp01-52 Grade 9 Core Answers | 9年级核心答案（第1-52页）
This revision guide covers the key answers and methods found on pages 1 to 52 of the Grade 9 Core Mathematics workbook. It is designed for students preparing for checkpoint tests, end-of-term exams or internal school assessments. Every section pairs a concise English explanation with a matching Chinese translation, so learners can check their working, confirm the final answer and avoid the mistakes that appear most often in this part of the book.
本复习指南涵盖9年级核心数学练习册第1至52页的关键答案与方法，适用于备考 checkpoint 测试、期末考试或校内测评的学生。每个小节都配有简洁的英文说明与对应的中文译文，帮助学习者核对解题过程、确认最终答案，并避免本书这一部分最常出现的错误。
1. Number and Place Value | 数与位值
a value of 4 000 and the 2 has a value of 2. To write a number in expanded form, split it into thousands, hundreds, tens and units: 4 752 = 4 000 + 700 + 50 + 2.
第1-8页重点考查位值、整数排序与四舍五入。数字 7 在 4 752 中的值为 700，因为它位于百位；数字 4 的值为 4 000，数字 2 的值为 2。要把一个数写成展开式，需将其拆分为千位、百位、十位和个位：4 752 = 4 000 + 700 + 50 + 2。
| Digits | 4 | 7 | 5 | 2 |
| Place value | thousands | hundreds | tens | units |
| Value | 4 000 | 700 | 50 | 2 |
A common answer on these pages is to round 38 476 to the nearest thousand, which gives 38 000 because the hundreds digit is 4, and therefore we round down. Rounding the same number to the nearest hundred gives 38 500 because the tens digit is 7, so we round up.
这几页中一个常见答案是：将 38 476 四舍五入到最接近的千位得 38 000，因为百位数字是 4，所以舍去；将同一个数四舍五入到最接近的百位得 38 500，因为十位数字是 7，所以进位。
38 476 → 38 000 (nearest 1 000) → 38 500 (nearest 100)
When ordering integers from smallest to largest, remember that negative numbers are always smaller than positive numbers, and that a number further from zero on the left is smaller. For example, -4, -1, 0, 3, 8 is the correct ascending order, while 8, 3, 0, -1, -4 is the correct descending order.
在将整数从小到大排序时，请记住负数永远小于正数，而且在数轴上越靠左（离 0 越远）的数越小。例如 -4、-1、0、3、8 是正确的升序排列，而 8、3、0、-1、-4 是正确的降序排列。
2. Fractions, Decimals and Percentages | 分数、小数与百分数
nator: 3 ÷ 8 = 0.375. To change a decimal to a percentage, multiply by 100, so 0.375 = 37.5%. Going the other way, 45% = 0.45 = 45/100 = 9/20 in its simplest form.
第9-16页涉及分数、小数与百分数之间的相互转换。把分数化成小数，只需用分子除以分母：3 ÷ 8 = 0.375。把小数化成百分数，乘以 100，所以 0.375 = 37.5%。反向转换时，45% = 0.45 = 45/100，约分后为最简分数 9/20。
8/12 + 3/12 = 11/12. For multiplication you simply multiply across: 2/3 × 3/5 = 6/15 = 2/5. For division, keep the first fraction, turn the division sign into a multiplication sign, and flip the second fraction: 3/4 ÷ 2/5 = 3/4 × 5/2 = 15/8 = 1⅞.
做分数加减法时，必须先通分。对于 2/3 + 1/4，最小公分母是 12，于是式子变为 8/12 + 3/12 = 11/12。做乘法时直接分子乘分子、分母乘分母：2/3 × 3/5 = 6/15 = 2/5。做除法时，保持第一个分数不变，把除号改成乘号，并将第二个分数取倒数：3/4 ÷ 2/5 = 3/4 × 5/2 = 15/8 = 1⅞。
ange questions use the same structure every time: change ÷ original × 100. If a coat costs £80 and is reduced to £68, the change is £12, so the percentage decrease is 12 ÷ 80 × 100 = 15% as compound interest, use the multiplier method rather than adding percentages separately.
百分数变化题每次都使用同样的结构：变化量 ÷ 原值 × 100。如果一件外套原价 80 英镑降到 68 英镑，变化量是 12 英镑，所以降幅为 12 ÷ 80 × 100 = 15%。对于复利等连续多次增长的问题，应使用乘数法，而不是把百分数逐个相加。
percentage change = change ÷ original × 100
3. Ratio, Proportion and Rates | 比、比例与速率
Pages 17-22 cover ratio notation and sharing in a given ratio. To simplify a ratio, divide both parts by their highest common factor: 24 : 36 becomes 2 : 3 after dividing both by 12. Always write ratios in their simplest whole-number form and keep the order of the parts the same as in the question.
第17-22页考查比的表示法以及按给定比进行分配。化简比时，用两数的最大公因数去除这两部分：24 : 36 同时除以 12 得到 2 : 3。比一定要写成最简整数比，并且各部分的顺序要与题目保持一致。
o find the total number of shares, divide the amount by that total, then multiply. Sharing £240 in the ratio 3 : 5 gives 3 + 5 = 8 shares, so one share is £240 ÷ 8 = £30, and the two amounts are £90 and £150. Checking that £90 + £150 = £240 confirms the answer.
按给定比分配金额时，先把各部分相加求出总份数，用总量除以总份数，再分别相乘。把 240 英镑按 3 : 5 分配，总份数为 3 + 5 = 8，所以一份是 240 ÷ 8 = 30 英镑，两部分分别是 90 英镑和 150 英镑。用 90 + 150 = 240 英镑检验即可确认答案。
nce ÷ time, which rearranges to distance = speed × time and time = distance ÷ speed. For direct proportion, both quantities increase together and the ratio y ÷ x stays constant; for inverse proportion, one increases while the other decreases and the product x × y stays constant.
速率比较的是两个不同单位，例如千米每小时或每千克的价格。速度、距离与时间由公式 速度 = 距离 ÷ 时间 联系，它可以变形为 距离 = 速度 × 时间 和 时间 = 距离 ÷ 速度。正比例中两个量同时增大，比值 y ÷ x 保持不变；反比例中一个量增大时另一个量减小，乘积 x × y 保持不变。
4. Indices, Roots and Standard Form | 指数、根式与标准形式
÷ aⁿ = aᵐ⁻ⁿ. A power raised to another power multiplies the indices, so (aᵐ)ⁿ = aᵐⁿ. Any non-zero number to the power zero equals 1, and a negative index means the reciprocal, so a⁻ⁿ = 1 ÷ aⁿ.
第23-28页考查指数法则。同底数幂相乘时指数相加，相除时指数相减：aᵐ × aⁿ = aᵐ⁺ⁿ，aᵐ ÷ aⁿ = aᵐ⁻ⁿ。幂的乘方要把指数相乘，即 (aᵐ)ⁿ = aᵐⁿ。任何非零数的零次幂都等于 1，负指数表示取倒数，即 a⁻ⁿ = 1 ÷ aⁿ。
A typical answer on these pages is 2⁵ = 32, √144 = 12 and ∛27 = 3. Fractional indices combine a root with a power: a1/2 is the square root of a, and 272/3 means the cube root of 27 squared, which is 3² = 9.
这几页的典型答案是 2⁵ = 32、√144 = 12 和 ∛27 = 3。分数指数把开方与乘方结合起来：a1/2 表示 a 的平方根，272/3 表示 27 的立方根再平方，即 3² = 9。
itten as 6.2 × 10⁻⁴ g. To multiply in standard form, multiply the front numbers and add the powers of ten, then adjust the answer so that the front number is again between 1 and 10.
标准形式把一个数写成 A × 10ⁿ，其中 1 ≤ A < 10，n 为整数。距离 3.4 × 10⁵ 千米等于 340 000 千米，而很小的质量 0.00062 克写成 6.2 × 10⁻⁴ 克。用标准形式做乘法时，先把前面的数字相乘、把 10 的指数相加，再调整结果，使前面的数字重新落在 1 与 10 之间。
(2 × 10³) × (3 × 10⁴) = 6 × 10⁷
5. Algebraic Expressions and Factorising | 代数式与因式分解
a bracket is multiplied by a negative number, every sign inside the bracket changes: -3(x - 4) = -3x + 12.
第29-34页重点考查化简代数式与展开括号。只有同类项才能合并，所以 5x + 3y - 2x + 7y 化简为 3x + 10y。当括号乘以负数时，括号内每一项的符号都要改变：-3(x - 4) = -3x + 12。
, giving x² - 2x - 15. Two special cases appear frequently: the perfect square (x + 4)² = x² + 8x + 16, and the difference of two squares x² - 49 = (x + 7)(x - 7).
展开两个括号时，要用第一个括号中的每一项去乘第二个括号中的每一项。对于 (x + 3)(x - 5)，四个乘积分别是 x²、-5x、3x 和 -15，得到 x² - 2x - 15。有两种特殊情况频繁出现：完全平方 (x + 4)² = x² + 8x + 16，以及平方差 x² - 49 = (x + 7)(x - 7)。
uadratic such as x² - 7x + 12, find two numbers that multiply to give 12 and add to give -7, namely -3 and -4, so the answer is (x - 3)(x - 4). Always expand mentally to check the middle term.
因式分解是展开的逆过程。先提取最大公因式：6x² + 9x = 3x(2x + 3)。分解形如 x² - 7x + 12 的二次式时，要找到两个数，它们相乘得 12、相加得 -7，即 -3 和 -4，因此答案是 (x - 3)(x - 4)。一定要在脑中展开检验一次中间项。
6. Linear Equations, Inequalities and Formulae | 一元一次方程、不等式与公式变形
to get 3x = 15, then divide by 3 to obtain x = 5. Always substitute your answer back into the original equation as a check.
第35-40页考查解方程。对等式一边所做的运算，必须同样施加于另一边。对于 3x + 7 = 22，两边同时减 7 得 3x = 15，再除以 3 得 x = 5。一定要把答案代回原方程检验。
x = 18 and x = 9. Equations with fractions should be cleared by multiplying every term by the common denominator.
当方程中含有括号时，先展开，或用括号前的系数去除两边。对于 5(x - 2) = 3x + 8，展开得 5x - 10 = 3x + 8，所以 2x = 18，x = 9。含分数的方程应通过每一项都乘以公分母来去分母。
Inequalities are solved in the same way, with one crucial exception: multiplying or dividing by a negative number reverses the inequality sign. So from -2x > 8 we get x < -4, and the solution is shown on a number line with an open circle at -4 and an arrow pointing left.
不等式的解法与之相同，但有一个关键例外：乘以或除以负数时，不等号方向要反转。因此由 -2x > 8 得到 x < -4，其解在数轴上用 -4 处的空心圆圈和向左的箭头表示。
e new subject appears twice, collect those terms on one side and factorise before dividing.
公式变形遵循同样的平衡原理。由 v = u + at，两边同时减 u 再除以 a，得 t = (v - u) ÷ a。如果待求字母在式子中出现两次，先把这些项集中到一边并提取公因式，再作除法。
7. Sequences and Patterns | 数列与规律
The nth term rule is found by comparing the sequence with the 4 times table, giving 4n - 1. The twentieth term is therefore 4 × 20 - 1 = 79.
第41-44页涉及数列。等差数列中每次加上同一个数；数列 3、7、11、15 的公差是 4。把该数列与 4 的乘法表比较，可得第 n 项的公式为 4n - 1。因此第 20 项是 4 × 20 - 1 = 79。
If the first differences are not constant but change by a fixed amount, the sequence is quadratic. For 2, 5, 10, 17, 26 the differences are 3, 5, 7, 9, so the second difference is 2. The coefficient of n² is half of the second difference, here 1, and comparing with n² gives the rule n² + 1.
如果一阶差不是常数，而是按固定数量变化，则该数列是二次数列。对于 2、5、10、17、26，差分别为 3、5、7、9，因此二阶差为 2。n² 的系数是二阶差的一半，此处为 1，再与 n² 比较即可得到公式 n² + 1。
Special sequences worth memorising include the square numbers 1, 4, 9, 16, 25; the triangular numbers 1, 3, 6, 10, 15; the cube numbers 1, 8, 27, 64, 125; and the Fibonacci sequence 1, 1, 2, 3, 5, 8, in which each term is the sum of the two before it.
值得记忆的特殊数列包括：平方数 1、4、9、16、25；三角数 1、3、6、10、15；立方数 1、8、27、64、125；以及斐波那契数列 1、1、2、3、5、8，其中每一项都是前两项之和。
8. Coordinates, Straight-Line Graphs and Simultaneous Equations | 坐标、直线图像与联立方程
= (y₂ - y₁) ÷ (x₂ - x₁). For the points (2, 5) and (4, 11), m = (11 - 5) ÷ (4 - 2) = 3, and substituting (2, 5) into y = mx + c gives 5 = 6 + c, so c = -1 and the equation is y = 3x - 1.
第45-48页把坐标几何与用图像法解联立方程结合起来。过两点的直线斜率为 m = (y₂ - y₁) ÷ (x₂ - x₁)。对于点 (2, 5) 与 (4, 11)，m = (11 - 5) ÷ (4 - 2) = 3，把 (2, 5) 代入 y = mx + c 得 5 = 6 + c，故 c = -1，直线方程为 y = 3x - 1。
Parallel lines have equal gradients, while perpendicular lines have gradients whose product is -1, for example 2 and -½. A line with a positive gradient rises from left to right, and a line with a negative gradient falls. The point where a graph crosses the y-axis is always the value of c.
平行线的斜率相等，而垂直线的斜率之积为 -1，例如 2 与 -½。斜率为正的直线从左向右上升，斜率为负的直线则下降。图像与 y 轴的交点纵坐标始终就是 c 的值。
. On a graph this answer is the coordinates of the point where the two lines cross, which is why the graphical and algebraic methods must always agree.
联立方程可以用消元法求解。已知 2x + y = 11 与 x - y = 1，两式相加消去 y，得 3x = 12，所以 x = 4、y = 3。在图像上，这个答案就是两条直线的交点坐标，这正是图像法与代数法结果必然一致的原因。
2x + y = 11 and x - y = 1 → x = 4, y = 3
9. Angles, Polygons and Circle Facts | 角度、多边形与圆的性质
Pages 49-50 revise angle facts. Angles on a straight line add to 180°, angles around a point add to 360°, and vertically opposite angles are equal. The interior angles of any triangle add to 180°, which means the angles of a quadrilateral add to 360° because it can be split into two triangles.
第49-50页复习角度知识。直线上的邻补角之和为 180°，围绕一点的角度之和为 360°，对顶角相等。任意三角形的内角和为 180°，因此四边形可以分成两个三角形，其内角和为 360°。
lygon all exterior angles are equal and add to 360°, so each exterior angle equals 360° ÷ n; for a regular dodecagon this is 360 ÷ 12 = 30°, and each interior angle is 180° - 30° = 150°.
对于任何有 n 条边的多边形，内角和为 (n -
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