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If x is numerically so small so that $x^2$ and higher powers of x can be neglected, then $\bigg(1+ \large\frac{2x}{3}\bigg)^{\large\frac{3}{2}}$$.(32+5x)^{\large\frac{-1}{5}}$ is approximately equal to :

\[\begin {array} {1 1} (1)\;\frac{32+31x}{64} & \quad (2)\;\frac{31+32x}{64} \\ (3)\;\frac{31-32x}{64} & \quad (4)\;\frac{1-2x}{64} \end {array}\]

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$ (1)\;\frac{32+31x}{64}$
answered Nov 7, 2013 by pady_1
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