数値と代数学
国際バカロレア · IB Diploma · Mathematics: Applications and Interpretation · SL トピック 1 5:07 英語ナレーション・英語＋中文字幕 burning-in
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Transcript
These are selected topics, not the whole syllabus, and every example is fictional.
这些是精选主题，不是完整大纲，所有例子都是虚构的。
The highest common factor takes the smallest shared prime powers; the lowest common multiple takes the largest.
最大公因数取共有质因数的最小幂；最小公倍数取所有质因数的最大幂。
A percentage change is one multiplier, and you reverse it by dividing.
百分比变化就是乘一个乘数，反推时用除法。
Arithmetic sequences add a common difference, geometric ones multiply by a common ratio.
等差数列加公差，等比数列乘公比。
A deposit stream follows the recurrence B n equals r times the last balance plus d.
定期存款满足递推：Bₙ = r × 上一期余额 + d。
A supplier packs 72 pencils and 90 pens into identical bags with nothing left over.
供应商把 72 支铅笔和 90 支钢笔装进相同的袋子，不能剩。
Split each number into primes: 72 is 2 cubed times 3 squared, and 90 is 2 times 3 squared times 5.
把每个数分解质因数：72 = 2³ × 3²，90 = 2 × 3² × 5。
The primes they share sit in the overlap: 2, 3 and 3, so the HCF is 18.
共有的质因数在重叠部分：2、3、3，所以最大公因数是 18。
That makes 18 bags of 4 pencils and 5 pens.
可装 18 袋，每袋 4 支铅笔和 5 支钢笔。
Multiply every prime in the diagram for the LCM: 360.
把图中所有质因数相乘得最小公倍数：360。
Now 84 pencils and 126 pens.
再看 84 支铅笔和 126 支钢笔。
84 is 2 squared times 3 times 7, and 126 is 2 times 3 squared times 7.
84 = 2² × 3 × 7，126 = 2 × 3² × 7。
The shared part is 2 times 3 times 7, which is 42, so 42 bags of 2 pencils and 3 pens.
共有部分是 2 × 3 × 7 = 42，所以装 42 袋，每袋 2 支铅笔和 3 支钢笔。
The LCM, 252, answers a different question: when repeated packs first match.
最小公倍数 252 回答的是另一个问题：重复包装何时第一次对齐。
And an HCF can never be bigger than the smaller number.
最大公因数永远不会大于较小的那个数。
A coat is reduced by 20 percent to 240 yuan.
一件外套降价 20% 后是 240 元。
The discount is of the original price, so the sale price is 0.8 times P.
折扣按原价计算，所以售价是 0.8P。
Divide: P is 240 over 0.8, which is 300.
相除：P = 240 ÷ 0.8 = 300。
Now raise the sale price by 20 percent: 240 times 1.2 is 288, not 300, because the second 20 percent is of the smaller amount.
再把售价提高 20%：240 × 1.2 = 288，不是 300，因为第二个 20% 是按较小的数计算的。
Proportion starts with a constant k found from a known pair.
比例先由已知的一对数求常数 k。
Suppose y is 6 when x is 2.
设 x = 2 时 y = 6。
If y is directly proportional to x, then y equals k x and k is 3: the line y equals 3 x.
若 y 与 x 成正比，则 y = kx，k = 3：直线 y = 3x。
If y is inversely proportional, then y equals k over x and k is 12: the curve y equals 12 over x.
若成反比，则 y = k/x，k = 12：曲线 y = 12/x。
Doubling x doubles one and halves the other.
x 加倍时，一个加倍，另一个减半。
Start at 5 and add 3 each time: 5, 8, 11.
从 5 开始每次加 3：5、8、11。
Plotted against n, equal steps of 3 put the terms on a straight line.
以 n 为横轴作图，相等的步长 3 使各项落在一条直线上。
The eighth term adds seven steps, not eight: 5 plus 7 times 3 is 26.
第 8 项加了 7 步而不是 8 步：5 + 7 × 3 = 26。
The sum of the first eight terms is 8 over 2 times 5 plus 26, which is 124.
前八项和是 8/2 × (5 + 26) = 124。
Now multiply instead: 12, 6, 3, and so on, with ratio a half.
改为相乘：12、6、3……公比 ½。
The running totals are 12, 18, 21, 22.5: each step closes half the remaining gap to 24.
累计和是 12、18、21、22.5：每一步都把到 24 的差距缩小一半。
So the sum to infinity is 12 over 1 minus a half, which is 24.
所以无穷和是 12 ÷ (1 − ½) = 24。
With a ratio of 2 the terms grow without limit and there is no sum to infinity.
公比为 2 时各项无限增大，没有无穷和。
Deposit 100 at the end of each period at 5 percent growth.
每期期末存 100，增长率 5%。
Each balance is 1.05 times the last, plus 100: 100, 205, 315.25.
每期余额是上期的 1.05 倍再加 100：100、205、315.25。
The closed form is 100 times 1.05 to the n minus 1, over 0.05.
通项是 100 × (1.05ⁿ − 1)/0.05。
B4 is 431.01, still below the target, and B5 is 552.56, so 5 periods first suffice.
B₄ = 431.01，仍低于目标；B₅ = 552.56，所以第一次够 5 期。
Paying at the start of each period earns one extra period of interest, so that line runs above.
期初存款多赚一期利息，所以那条线在上方。
Unroll the recurrence.
展开递推。
After three periods the first deposit has grown twice, the second once and the last not at all: 100 times 1.05 squared plus 1.05 plus 1, which is 315.25.
三期后，第一笔存款增长了两次，第二笔一次，最后一笔还没有增长：100 × (1.05² + 1.05 + 1) = 315.25。
That is a geometric series, so B n is 100 times 1.05 to the n minus 1, over 0.05.
这是等比数列，所以 Bₙ = 100 × (1.05ⁿ − 1)/0.05。
Deposits at the start of each period all earn one more period: C3 is 331.01, not 315.25.
期初存款每笔都多一期：C₃ = 331.01，而不是 315.25。
Check yourself.
自我检测。
The HCF of 20 and 28 is 4, and their LCM is 140.
20 和 28 的最大公因数是 4，最小公倍数是 140。
400 down 10 percent and up 10 percent ends at 396, not 400.
400 先降 10% 再升 10%，结果是 396，不是 400。
With first term 9 and difference 2, the 6th term is 9 plus 5 twos: 19.
首项 9、公差 2，第 6 项是 9 + 5 × 2 = 19。
B two is 1.05 times 100, plus 100: 205.
B₂ = 1.05 × 100 + 100 = 205。