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If $|x|$ is so small that $x^3$ and higher powers of $x$ can be neglected, then an approximate value of $\frac{1}{\sqrt{4-x}(2+x)^3}$ is
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The correct answer is:
$\frac{1}{16}\left(1-\frac{11 x}{8}+\frac{171}{128} x^2\right)$
$\frac{1}{\sqrt{4-x}(2+x)^3}=(4-x)^{-1 / 2} \cdot(2+x)^{-3}$
$\begin{aligned} & =4^{-1 / 2} \cdot 2^{-3}\left(1-\frac{x}{4}\right)^{-1 / 2}\left(1+\frac{x}{2}\right)^{-3} \\ & =\frac{i}{16}\left(1+\frac{1}{2} \cdot \frac{x}{4}+\frac{1}{2}\left(\frac{3}{2}\right) \frac{x^2}{16}+. .\right)\left(1+(-3)\left(\frac{x}{2}\right)+3(4)\left(\frac{x}{2}\right)^2+. .\right) \\ & =\frac{1}{16}\left(1-\frac{3 x}{2}+\frac{x}{8}+6 x^2-\frac{3}{16} x^2+\frac{3}{64} x^2\right) \\ & =\frac{1}{16}\left(1-\frac{11 x}{8}+\frac{171}{128} x^2\right) .\end{aligned}$
$\begin{aligned} & =4^{-1 / 2} \cdot 2^{-3}\left(1-\frac{x}{4}\right)^{-1 / 2}\left(1+\frac{x}{2}\right)^{-3} \\ & =\frac{i}{16}\left(1+\frac{1}{2} \cdot \frac{x}{4}+\frac{1}{2}\left(\frac{3}{2}\right) \frac{x^2}{16}+. .\right)\left(1+(-3)\left(\frac{x}{2}\right)+3(4)\left(\frac{x}{2}\right)^2+. .\right) \\ & =\frac{1}{16}\left(1-\frac{3 x}{2}+\frac{x}{8}+6 x^2-\frac{3}{16} x^2+\frac{3}{64} x^2\right) \\ & =\frac{1}{16}\left(1-\frac{11 x}{8}+\frac{171}{128} x^2\right) .\end{aligned}$
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