题目

Problem

Expand $(9-3x)^{\frac12}$ in ascending powers of $x$ , up to and including the term in $x^2$ , simplifying the coefficients.

(4)

题目中文翻译

将 $(9-3x)^{\frac12}$ 按 $x$ 的升幂展开，直至并包括 $x^2$ 项，并化简各项系数。

(4)

原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

解法一

思路

展开

先提出平方根内的九，标准二项展开中代入 $-x/3$ ，最后别忘记乘三。

答题过程

展开

Write

(9-3x)^{1/2}=3\bigg(1-\frac x3\bigg)^{1/2}.

Using the binomial expansion,

\begin{align*} \bigg(1-\frac x3\bigg)^{1/2} =&\,1+\frac12\bigg(-\frac x3\bigg)\\[4mm] &\,\hspace{2pt}+\frac{(1/2)(-1/2)}{2!}\bigg(-\frac x3\bigg)^2+\cdots\\[4mm] =&\,1-\frac x6-\frac{x^2}{72}+\cdots. \end{align*}

Multiplying by $3$ gives

\boxed{(9-3x)^{1/2}=3-\frac x2-\frac{x^2}{24}+\cdots}.

CIE 9709 2024 November Paper 32 Q1

题目

解答

解法一

思路

答题过程