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If $x$ is so small that $x^2$ and higher powers of $x$ may be neglected, then the approximate value of $\frac{\left(1+\frac{2}{3} x\right)^{-3}(1-15 x)^{-1 / 5}}{(2-3 x)^4}$
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Verified Answer
The correct answer is:
$\frac{1}{16}(1+7 x)$
We have,
$$
\begin{aligned}
& \frac{\left(1+\frac{2}{3} x\right)^{-3}(1-15 x)^{-1 / 5}}{(2-3 x)^4} \\
= & \frac{\left(1+\frac{2}{3} x\right)^{-3}(1-15 x)^{-1 / 5}}{2^4\left(1-\frac{3}{2} x\right)^4} \\
= & \frac{\left(1+\frac{2}{3} x\right)^{-3}(1-15 x)^{-1 / 5}\left(1-\frac{3}{2} x\right)^{-4}}{16} \\
= & \frac{1}{16}(1-2 x)(1+3 x)(1+6 x) \\
= & \frac{1}{16}(1+x)(1+6 x) \\
= & \frac{1}{16}(1+7 x)
\end{aligned}
$$
$$
\begin{aligned}
& \frac{\left(1+\frac{2}{3} x\right)^{-3}(1-15 x)^{-1 / 5}}{(2-3 x)^4} \\
= & \frac{\left(1+\frac{2}{3} x\right)^{-3}(1-15 x)^{-1 / 5}}{2^4\left(1-\frac{3}{2} x\right)^4} \\
= & \frac{\left(1+\frac{2}{3} x\right)^{-3}(1-15 x)^{-1 / 5}\left(1-\frac{3}{2} x\right)^{-4}}{16} \\
= & \frac{1}{16}(1-2 x)(1+3 x)(1+6 x) \\
= & \frac{1}{16}(1+x)(1+6 x) \\
= & \frac{1}{16}(1+7 x)
\end{aligned}
$$
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