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If $x$ is small, so that $x^2$ and higher powers can be neglected, then the approximate value for $\frac{(1-2 x)^{-1}(1-3 x)^{-2}}{(1-4 x)^{-3}}$ is
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Verified Answer
The correct answer is:
$1-2 x$
Given expression is
$$
\begin{aligned}
& E=\frac{(1-2 x)^{-1}(1-3 x)^{-2}}{(1-4 x)^{-3}} \\
&=\frac{\left(1+2 x+2 x^2+\ldots\right)(1+6 x+\ldots)}{(1+12 x+\ldots)} \\
&=\frac{(1+2 x+6 x+\ldots)}{(1+12 x)} \\
&=(1+8 x)(1+12 x)^{-1} \\
&=(1+8 x)(1-12 x \ldots) \\
&=(1+8 x-12 x \ldots) \\
&=(1-4 x) \\
& \quad(\because \text { neglecting the higher term })
\end{aligned}
$$
( $\because$ neglecting the higher term)
$$
\begin{aligned}
& E=\frac{(1-2 x)^{-1}(1-3 x)^{-2}}{(1-4 x)^{-3}} \\
&=\frac{\left(1+2 x+2 x^2+\ldots\right)(1+6 x+\ldots)}{(1+12 x+\ldots)} \\
&=\frac{(1+2 x+6 x+\ldots)}{(1+12 x)} \\
&=(1+8 x)(1+12 x)^{-1} \\
&=(1+8 x)(1-12 x \ldots) \\
&=(1+8 x-12 x \ldots) \\
&=(1-4 x) \\
& \quad(\because \text { neglecting the higher term })
\end{aligned}
$$
( $\because$ neglecting the higher term)
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