题目
Problem
Expand 1 + 2 x 1 − 2 x \sqrt{\dfrac{1+2x}{1-2x}} 1 − 2 x 1 + 2 x x x x x 2 x^2 x 2
(5)
题目中文翻译
将 1 + 2 x 1 − 2 x \sqrt{\dfrac{1+2x}{1-2x}} 1 − 2 x 1 + 2 x x x x x 2 x^2 x 2
(5)
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
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先把根式拆成两个非整数幂，分别展开到二次项，再收集乘积中的同次项。
答题过程
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Using the binomial expansion,
( 1 + 2 x ) 1 / 2 = 1 + 1 2 ( 2 x ) + ( 1 / 2 ) ( − 1 / 2 ) 2 ( 2 x ) 2 + ⋯ = 1 + x − 1 2 x 2 + ⋯ , \begin{align*}
(1+2x)^{1/2}
=&\,1+\frac12(2x)
+\frac{(1/2)(-1/2)}2(2x)^2+\cdots\\[4mm]
=&\,1+x-\frac12x^2+\cdots,
\end{align*} ( 1 + 2 x ) 1/2 = = 1 + 2 1 ( 2 x ) + 2 ( 1/2 ) ( − 1/2 ) ( 2 x ) 2 + ⋯ 1 + x − 2 1 x 2 + ⋯ , and
( 1 − 2 x ) − 1 / 2 = 1 − 1 2 ( − 2 x ) + ( − 1 / 2 ) ( − 3 / 2 ) 2 ( − 2 x ) 2 + ⋯ = 1 + x + 3 2 x 2 + ⋯ . \begin{align*}
(1-2x)^{-1/2}
=&\,1-\frac12(-2x)
+\frac{(-1/2)(-3/2)}2(-2x)^2+\cdots\\[4mm]
=&\,1+x+\frac32x^2+\cdots.
\end{align*} ( 1 − 2 x ) − 1/2 = = 1 − 2 1 ( − 2 x ) + 2 ( − 1/2 ) ( − 3/2 ) ( − 2 x ) 2 + ⋯ 1 + x + 2 3 x 2 + ⋯ . Multiplying,
1 + 2 x 1 − 2 x = ( 1 + x − 1 2 x 2 + ⋯ ) ( 1 + x + 3 2 x 2 + ⋯ ) = 1 + 2 x + ( 3 2 + 1 − 1 2 ) x 2 + ⋯ = 1 + 2 x + 2 x 2 + ⋯ . \begin{align*}
\sqrt{\frac{1+2x}{1-2x}}
=&\,(1+x-\tfrac12x^2+\cdots)
(1+x+\tfrac32x^2+\cdots)\\[4mm]
=&\,1+2x+\bigg(\frac32+1-\frac12\bigg)x^2+\cdots\\[4mm]
=&\,\boxed{1+2x+2x^2+\cdots}.
\end{align*} 1 − 2 x 1 + 2 x = = = ( 1 + x − 2 1 x 2 + ⋯ ) ( 1 + x + 2 3 x 2 + ⋯ ) 1 + 2 x + ( 2 3 + 1 − 2 1 ) x 2 + ⋯ 1 + 2 x + 2 x 2 + ⋯ . The expansion is valid for ∣ x ∣ < 1 / 2 |x|<1/2 ∣ x ∣ < 1/2
解法二
思路
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把分子、分母同时乘上 1 + 2 x 1+2x 1 + 2 x x 2 x^2 x 2
答题过程
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For ∣ x ∣ < 1 / 2 |x|<1/2 ∣ x ∣ < 1/2 1 + 2 x > 0 1+2x>0 1 + 2 x > 0
1 + 2 x 1 − 2 x = 1 + 2 x 1 − 4 x 2 = ( 1 + 2 x ) ( 1 − 4 x 2 ) − 1 / 2 . \begin{align*}
\sqrt{\frac{1+2x}{1-2x}}
=&\,\frac{1+2x}{\sqrt{1-4x^2}}\\[4mm]
=&\,(1+2x)(1-4x^2)^{-1/2}.
\end{align*} 1 − 2 x 1 + 2 x = = 1 − 4 x 2 1 + 2 x ( 1 + 2 x ) ( 1 − 4 x 2 ) − 1/2 . Using the binomial expansion,
( 1 − 4 x 2 ) − 1 / 2 = 1 + ( − 1 2 ) ( − 4 x 2 ) + ⋯ = 1 + 2 x 2 + ⋯ . (1-4x^2)^{-1/2}
=1+\bigg(-\frac12\bigg)(-4x^2)+\cdots
=1+2x^2+\cdots. ( 1 − 4 x 2 ) − 1/2 = 1 + ( − 2 1 ) ( − 4 x 2 ) + ⋯ = 1 + 2 x 2 + ⋯ . Therefore, up to and including x 2 x^2 x 2
1 + 2 x 1 − 2 x = ( 1 + 2 x ) ( 1 + 2 x 2 + ⋯ ) = 1 + 2 x + 2 x 2 + ⋯ . \begin{align*}
\sqrt{\frac{1+2x}{1-2x}}
=&\,(1+2x)(1+2x^2+\cdots)\\[4mm]
=&\,\boxed{1+2x+2x^2+\cdots}.
\end{align*} 1 − 2 x 1 + 2 x = = ( 1 + 2 x ) ( 1 + 2 x 2 + ⋯ ) 1 + 2 x + 2 x 2 + ⋯ . 