OCR数学 代数 因式分解 方程 不等式 数列
OCR GCSE 数学 J560 的代数在官方考纲里分成六块:代数式、公式、方程、不等式、函数语言和数列。考试是三张卷子,每张 100 分、1 小时 30 分,各占 33⅓%,其中 Foundation 的 Paper 02 和 Higher 的 Paper 05 不允许使用计算器。这意味着不少代数题要在没有计算器的情况下手算到底,根号与平方得练到不假思索。下面按学生真正会卡住的地方拆开讲,每个小节配一个能算完的例子。
Algebra on OCR GCSE Mathematics (J560) is listed in the specification under six headings: expressions, formulae, equations, inequalities, the language of functions and sequences. Candidates sit three papers of 100 marks and 1 hour 30 minutes each, weighted at 33⅓% apiece, and one paper per tier is non-calculator (J560/02 at Foundation, J560/05 at Higher), so much of the algebra has to be finished by hand. The sections below follow the points where marks are actually lost, and each one carries a worked example you can check.
同类项合并与代数式的书写规范
代数式的写法有几条硬规矩,乘号通常省略,3 乘 a 写成 3a,a 乘 b 写成 ab,除号换成分数线,a 除以 4 写成 a/4。合并同类项只看字母和指数是否相同,3x 与 5x 算一类,3x 与 3x^2 不算。很多学生以为 3x + 4y 还能再合并,实际上字母不同就只能停在这里。化简 5a - 2a + 7b - b 得到 3a + 6b,中间那一步是系数相减,不是把字母消掉。另一处丢分在负号,-2(x - 5) 展开是 -2x + 10,括号里第二项要变号。指数相同的判断靠眼睛,4x^2 与 x^2 是同类,4x^2 与 x^3 不是。写答案时把 3 乘 x 改回 3x,这种混写容易被当成笔误。同类项没合并完就往下算,后面每一步都带着多余的东西。
In algebra the notation itself is marked: 3 times a is written 3a, a times b is written ab, and a divided by 4 becomes a/4, so a stray multiplication sign can cost a mark even when the value is right. Like terms must match in both letter and power, which is why 3x and 5x combine but 3x and 3x^2 do not; 3x + 4y simply stays as it is. Simplifying 5a - 2a + 7b - b gives 3a + 6b, and the step that goes wrong most often is the negative bracket, where -2(x - 5) expands to -2x + 10. Examiners see the missing sign change every June.
展开双重括号与提取完整公因式
展开 (x + 3)(x - 5) 要把每一项都乘一遍。第一次乘得 x^2 - 5x,第二次乘得 3x - 15,合起来是 x^2 - 2x - 15。常见的错误是只乘首项和末项,写成 x^2 - 15,中间的 -2x 整段丢掉。反向的提取公因式要先找最大公因数,6x^2 + 9x 提出 3x 得 3x(2x + 3)。阅卷时看到只提一半的答案,比如 6x^2 + 9x 写成 3(2x^2 + 3x),通常不给满分,因为括号里还能继续提 x。检验用代数是三十秒的事,取 x = 1 时两个版本都该等于 -16。括号前是负号时全部项都要变号,这一条在双重括号里同样成立。展开与因式分解是互为反向的动作,一项一项核对比重做一遍省时间。
Expanding two brackets means every term in the first must meet every term in the second: (x + 3)(x - 5) gives x^2 - 5x + 3x - 15, and the tidy answer is x^2 - 2x - 15. A very common slip is multiplying only the first and last terms, which produces x^2 - 15 and throws away the whole middle term. Factorising runs the other way, and the highest common factor has to come out completely, so 6x^2 + 9x becomes 3x(2x + 3) rather than 3(2x^2 + 3x). Substitution is a quick self-check: put x = 1 into both versions and you should get -16 either way.
二次三项式因式分解与平方差识别
x^2 - 7x + 12 的因式分解要找两个数,乘积是 12、和是 -7,这对数是 -3 和 -4,于是写成 (x - 3)(x - 4)。符号的判断靠和与积,常数项为正、一次项为负,两个数就都是负数。平方差有固定形式,x^2 - 49 直接写 (x - 7)(x + 7),9x^2 - 25 写成 (3x - 5)(3x + 5)。学生最常把 x^2 - 49 当成不能再分解,或者在 9x^2 - 25 里漏掉系数 3。分解完把括号展开回来核对一遍,这一步在非计算器卷上比重新想一遍快得多。代入检验同样好用,把 x = 3 代进 (x - 3)(x - 4) 应该得 0。常数项为正而一次项也为正时,两个数都取正,四种符号组合可以这样推出来。题目要求写成因式相乘的形式时,别把答案再展开回去。
For x^2 - 7x + 12 the two numbers must multiply to 12 and add to -7, so they are -3 and -4 and the factorised form is (x - 3)(x - 4). The signs follow from the sum and the product, which is why a positive constant with a negative x term forces two negative numbers. A difference of two squares is recognised by shape rather than by trial: x^2 - 49 is (x - 7)(x + 7) and 9x^2 - 25 is (3x - 5)(3x + 5). Leaving x^2 - 49 unfactorised, or dropping the 3 in 9x^2 - 25, accounts for a lot of lost marks at Higher tier.
一元一次方程去分母与两边都有未知数
解 5(x - 2) = 3x + 8 先把括号展开,得到 5x - 10 = 3x + 8。两边同减 3x 得 2x - 10 = 8,再加 10 得 2x = 18,最后 x = 9。分数系数的方程先去分母,(x + 1)/2 + (x - 2)/3 = 4 两边乘 6,变成 3(x + 1) + 2(x - 2) = 24。把上式化简得 5x - 1 = 24,所以 x = 5。这一步最容易错在只乘了分子,或者漏乘等号右边的 4。去分母时括号不能省,3(x + 1) 里的 x + 1 是一个整体,漏掉括号就是整题错。两边都有未知数时先移项再合并,移项的本质是等式两边同加或同减。把答案代回原方程检验是三十秒的事,很多学生直接跳过,带着错答案往下做。
Start 5(x - 2) = 3x + 8 by expanding to 5x - 10 = 3x + 8, then subtract 3x from both sides to reach 2x - 10 = 8 and finish with x = 9. When fractions appear, clear the denominators first: multiplying (x + 1)/2 + (x - 2)/3 = 4 through by 6 gives 3(x + 1) + 2(x - 2) = 24 and then 5x - 1 = 24, so x = 5. The failure here is usually a term that never got multiplied, most often the number on the right of the equals sign. Substituting the answer back takes half a minute and catches nearly all of these errors.
公式变形:把指定字母放到等号左边
公式变形的目标是把某个字母单独留在等号一边。以 2y = 3x - 5 为例,要让 x 成为主语,先把 -5 移过去得 3x = 2y + 5,再两边除以 3 得 x = (2y + 5)/3。顺序上先处理加减、再处理乘除,和解放程一致。括号里的字母要先展开或者先整体提出来,a(x + b) = c 变形第一步就是两边除以 a。常见的错误是整段除以 3 时只除了第一项,写成 x = 2y + 5/3,这种答案在变形题里几乎不给分。分母上带字母的式子要整块乘过去,乘的是分母而不是分子。变形完成把结果代回原式,两边应该还原成同一个式子。题目要求写成 x 在等号左边的形式时,右边不要再留括号。
Rearranging a formula is about isolating one letter while everything else stays put. Take 2y = 3x - 5 and make x the subject: move the 5 across to get 3x = 2y + 5, then divide the whole expression by 3 so that x = (2y + 5)/3. Undo addition and subtraction before multiplication and division, and if the target sits inside a bracket, deal with the bracket first. Dividing only the first term, which turns the answer into 2y + 5/3, gives a wrong result that scores nothing.
二元一次方程组:加减消元与代入
二元一次方程组的两条路是加减消元和代入。看 2x + 3y = 12 与 4x - 3y = 6,两式的 y 系数一正一负,直接相加就消掉 y,得 6x = 18。于是 x = 3,代回第一式得 6 + 3y = 12,y = 2。系数不成倍数时先给一条式子乘一个数,让某个字母的系数对齐。答案要成对写出来,只写 x = 3 会被扣掉一半的分。检验时两个方程都代一遍,只代一条容易放过符号错误。草稿纸上画两条直线更快,交点的坐标就是方程组的解。解出负值不代表算错,代回去成立就是对的。
Two routes cover simultaneous equations: elimination and substitution. With 2x + 3y = 12 and 4x - 3y = 6 the y terms already cancel when the equations are added, giving 6x = 18 and x = 3, and back-substitution then gives y = 2. If the coefficients do not match, scale one equation first so that a letter lines up. Write the answer as a pair, because a lone x = 3 is incomplete, and check in both original equations rather than just one.
二次方程:因式分解优先于求根公式
二次方程先试因式分解,分解不出来再上公式。x^2 + 2x - 8 = 0 找乘积 -8、和 2 的两个数,是 -2 和 4,于是 (x + 4)(x - 2) = 0,解得 x = -4 或 x = 2。分解不动的用求根公式 x = (-b ± sqrt(b^2 - 4ac))/(2a)。以 2x^2 - 5x - 3 = 0 为例,a = 2、b = -5、c = -3,判别式是 25 + 24 = 49,开方得 7。代进公式得 x = (5 ± 7)/4,两个根是 3 和 -0.5。很多学生以为公式里的 b 直接代正数,负号得跟着一起进括号。两个根都要写,只写一个通常算答案不全。判别式为负时没有实根,这一句在 Higher 卷上常作为解释题出现。
Always try factorising a quadratic before reaching for the formula. For x^2 + 2x - 8 = 0 the pair -2 and 4 gives (x + 4)(x - 2) = 0, so x = -4 or x = 2. When the expression will not factorise, use x = (-b ± sqrt(b^2 - 4ac))/(2a); with 2x^2 - 5x - 3 = 0 the discriminant is 25 + 24 = 49, its square root is 7, and x = (5 ± 7)/4 gives 3 and -0.5. Keep the minus sign attached to b when substituting, and always give both roots.
一元一次不等式与整数解计数
解不等式 4x - 7 < 3x + 2,两边减 3x 得 x - 7 < 2,所以 x < 9。规则与方程相同,唯一的分叉是两边乘或除负数时不等号方向要翻转。双不等式 -3 ≤ 2x + 1 < 7 要三边一起动,先减 1 得 -4 ≤ 2x < 6,再除以 2 得 -2 ≤ x < 3。整数解是 -2、-1、0、1、2 共五个。常见的错误是数轴上把端点画成实心,x < 9 那个 9 要空心,因为 9 本身不满足条件。题目问整数解时先写不等式,再从数轴上一个一个点,比心算可靠。端点带等号时画实心,不带等号时画空心,两种记号别混。答案用不等式写比用文字描述清楚,阅卷时也更容易找到得分点。
Solving 4x - 7 < 3x + 2 gives x < 9 in exactly the same way as an equation, with one difference: multiplying or dividing by a negative number flips the inequality sign. A double inequality such as -3 ≤ 2x + 1 < 7 has to be worked on all three parts, so subtracting 1 gives -4 ≤ 2x < 6 and halving gives -2 ≤ x < 3. The integer solutions are -2, -1, 0, 1 and 2, five in total. On a number line the end point of x < 9 is open, since 9 itself does not satisfy the statement.
数列第 n 项:等差与二阶差分
等差数列 5、8、11、14 的公差是 3,第 n 项就含 3n。用 n = 1 校准得 3n + 2,验算 n = 2 得 8,对上了。差分为常数时是线性数列,二阶差分恒定则对应 n^2 项。看 2、7、14、23,一阶差分是 5、7、9,二阶差分恒为 2,说明含 n^2,待定系数算下来第 n 项是 n^2 + 2n - 1。学生最常把 n 当成项的值,或者忘了 n 只能取正整数。题目问某一项等于多少时,把通项设成那个数解方程,解出非整数就说明该项不存在。求第 50 项直接代 n = 50,不需要把前 50 项全写出来。通项写好后用两个相邻项各验一次,错一个符号就能立刻看出来。
The sequence 5, 8, 11, 14 rises by 3 each time, so its nth term starts with 3n, and calibrating on n = 1 gives 3n + 2, which checks out at n = 2. A constant first difference means a linear nth term, while a constant second difference means a quadratic one. For 2, 7, 14, 23 the first differences are 5, 7 and 9 and the second difference is 2, so the rule contains n^2 and works out as n^2 + 2n - 1. Remember that n counts positions and only takes positive whole numbers.
函数记号 f(x) 与反函数、复合顺序
函数记号 f(x) = 2x + 1 表示把输入代进规则,f(3) 就是把 3 放进去,得 7。复合函数看顺序,g(x) = x^2 时 fg(3) 是先算 g(3) = 9,再算 f(9) = 19,反过来 gf(3) 是 g(7) = 49。反函数 f^-1 的做法是把 y = 2x + 1 变形回 x,得 x = (y - 1)/2。把字母换回来写成 f^-1(x) = (x - 1)/2,记号就完整了。常见的错误是把 f^-1 当成 1/f(x),这两个完全不是一回事。检验反函数的方法是两个方向各代一次,f^-1(f(3)) 应该回到 3。定义域的限制在 Higher 卷里偶尔出现,分母不能为零是一条常见条件。函数题先把输入找清楚,再套规则,顺序弄反答案就换了一个。
Writing f(x) = 2x + 1 means the rule is applied to whatever goes in the bracket, so f(3) = 7. Composite functions depend on order: with g(x) = x^2, fg(3) applies g first and then f, giving 19, whereas gf(3) gives 49. An inverse function is found by rearranging y = 2x + 1 into x = (y - 1)/2 and relabelling, so f^-1(x) = (x - 1)/2. Never read f^-1 as 1/f(x); the two are unrelated, and checking that f^-1(f(3)) returns 3 settles the question.
把文字情境翻译成代数式
文字题的第一步是给未知量起名字,通常写 let the width be x。一个长方形的长比宽多 3 cm,周长 26 cm,于是一圈写成 2(x + x + 3) = 26。化简得 4x + 6 = 26,x = 5,长是 8 cm,面积 40 cm^2。名字起得含糊是这类题的主要卡点,两个量混在一个字母里,方程就写不下去。单位要跟着数字走,最后一行答面积时写 cm^2 而不是 cm。题目问的是哪个量就答哪个量,算出 x 只是中间结果。年龄、票价这类题先画一条时间线或者列一张小表,关系会清楚很多。建模题的过程分往往比答案分多,式子写清楚比算得快更值钱。
Word problems get easier once every unknown has a name, usually let the width be x. A rectangle 3 cm longer than it is wide with a perimeter of 26 cm gives 2(x + x + 3) = 26, then 4x + 6 = 26, so the width is 5 cm, the length is 8 cm and the area is 40 cm^2. Vague naming is where these questions stall, because two quantities sharing one letter cannot be separated later. Units belong to the final answer, which is cm^2 for area and not cm.
三张卷子的代数分布与步骤分
代数题不是只出现在一张卷子上,OCR 官方考纲写明主题可以出现在任何一张卷,所以同样的技巧会考三次。每张 100 分、1 小时 30 分,Foundation 的 Paper 02 与 Higher 的 Paper 05 不带计算器。这两张卷子里的根号、平方和分数要练到手算,算错一步后面全歪。步骤分是 OCR 明确给的,官方说明提到每张卷设 100 分为的是让学生在中间步骤上拿到方法分。算式要写下来而不是只在心里算,阅卷时最常看到的是答案对却没有过程。遇到稍难的题没有过程就一分不给,宁可多写一行。把要用的公式先抄在答题区边上,边算边代,比最后回头补过程稳。三张卷子的时间都是 90 分钟,代数题分配 25 到 30 分钟比较合理。
Algebra is not confined to one paper, since the specification states that topics may be assessed on any of them, so the same skills surface three times. Each paper carries 100 marks in 1 hour 30 minutes, and the non-calculator papers (J560/02 at Foundation, J560/05 at Higher) demand fluency with roots, squares and fractions done by hand. OCR awards method marks throughout, which is why each paper is worth 100 marks rather than a smaller total: a bare answer with no working shows nothing of the method. Write the algebra down line by line, and copy any formula you need into the working space before starting.
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