Assuming to be small that $x^{2}$ and higher powers of $x$ can be neglected, then $\frac{\left( 1 + \frac{3}{4}x \right)^{- 4}(16 - 3x)^{\frac{1}{2}}}{{(8 + x)}^{\frac{2}{3}}}$ is approximately equal to

Question

Dr. Abubacker · Accepted Answer

(b)
$\frac{\left( 1 + \frac{3}{4}x \right)^{- 4}(16)^{\frac{1}{2}}\left( 1 - \frac{3}{16}x \right)^{\frac{1}{2}}}{{(8)}^{\frac{2}{3}}\left( 1 + \frac{x}{8} \right)^{\frac{2}{3}}}$
$= \left( 1 + \frac{3x}{4} \right)^{- 4}\left( 1 - \frac{3x}{16} \right)^{\frac{1}{2}}\left( 1 + \frac{x}{8} \right)^{- \frac{2}{3}}$
$= \left( 1 + ( - 4)\frac{3}{4}x \right)\left( 1 - \left( \frac{1}{2} \right)\frac{3x}{16} \right)\left( 1 + \left( - \frac{2}{3} \right)\frac{x}{8} \right)$
$= (1 - 3x)\left( 1 - \frac{3}{32}x \right)\left( 1 - \frac{x}{12} \right)$
$= 1 - \frac{305}{96}x$ (On neglecting $x^{2}$ and higher powers of$\ x$)

Q. Assuming to be small that $x^{2}$ and higher powers of $x$ can be neglected, then $\frac{\left( 1 + \frac{3}{4}x \right)^{- 4}(16 - 3x)^{\frac{1}{2}}}{{(8 + x)}^{\frac{2}{3}}}$ is approximately equal to

Explanation

Reviewed by Dr. Abubacker

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