If |x| is so small that x^3 and higher powers | vedclass

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(D) We have $\frac{1}{\sqrt{4-x}(2+x)^3} = (4-x)^{-1/2} (2+x)^{-3}$.$= 4^{-1/2} \left(1-\frac{x}{4}\right)^{-1/2} \cdot 2^{-3} \left(1+\frac{x}{2}\rig

Vedclass · Accepted Answer

$\frac{1}{16}\left(1-\frac{11 x}{8}+\frac{171}{128} x^2\right)$

Vedclass · Answer

(D) We have $\frac{1}{\sqrt{4-x}(2+x)^3} = (4-x)^{-1/2} (2+x)^{-3}$.$= 4^{-1/2} \left(1-\frac{x}{4}\right)^{-1/2} \cdot 2^{-3} \left(1+\frac{x}{2}\right)^{-3}$$= \frac{1}{2} \cdot \frac{1}{8} \left(1 - \frac{1}{2}\left(-\frac{x}{4}\right) + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2} \left(-\frac{x}{4}\right)^2\right) \left(1 + (-3)\left(\frac{x}{2}\right) + \frac{(-3)(-4)}{2} \left(\frac{x}{2}\right)^2\right)$$= \frac{1}{16} \left(1 + \frac{x}{8} + \frac{3}{128} x^2\right) \left(1 - \frac{3x}{2} + \frac{3x^2}{2}\right)$$= \frac{1}{16} \left(1 - \frac{3x}{2} + \frac{3x^2}{2} + \frac{x}{8} - \frac{3x^2}{16} + \frac{3x^2}{128}\right)$$= \frac{1}{16} \left(1 - \frac{11x}{8} + \frac{171x^2}{128}\right)$.

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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