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If $\mathrm{x}$ is so small that $\mathrm{x}^3$ and higher powers of $\mathrm{x}$ may be neglected, then $\frac{(1+x)^{3 / 2}-\left(1+\frac{1}{2} x\right)^3}{(1-x)^{1 / 2}}$
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Verified Answer
The correct answer is:
$-\frac{3}{8} x^2$
$-\frac{3}{8} x^2$
$$
\begin{aligned}
& (1-x)^{1 / 2}\left[1+\frac{3}{2} x+\frac{3}{2}\left(\frac{3}{2}-1\right) x^2-1-3\left(\frac{1}{2} x\right)-3(2)\left(\frac{1}{2} x\right)^2\right] \\
& =(1-x)^{1 / 2}\left[-\frac{3}{8} x^2\right]=-\frac{3}{8} x^2
\end{aligned}
$$
\begin{aligned}
& (1-x)^{1 / 2}\left[1+\frac{3}{2} x+\frac{3}{2}\left(\frac{3}{2}-1\right) x^2-1-3\left(\frac{1}{2} x\right)-3(2)\left(\frac{1}{2} x\right)^2\right] \\
& =(1-x)^{1 / 2}\left[-\frac{3}{8} x^2\right]=-\frac{3}{8} x^2
\end{aligned}
$$
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