If x is so small that x^2 and higher powers o | vedclass

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(D) Given expression: $E = \frac{(1+\frac{2}{3}x)^{-3}(1-15x)^{-1/5}}{(2-3x)^4}$ Neglecting $x^2$ and higher powers,we use the binomial approximation

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$\frac{1}{16}(1+7x)$

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(D) Given expression: $E = \frac{(1+\frac{2}{3}x)^{-3}(1-15x)^{-1/5}}{(2-3x)^4}$ Neglecting $x^2$ and higher powers,we use the binomial approximation $(1+nx) \approx 1+nx$. $E = \frac{(1+\frac{2}{3}x)^{-3}(1-15x)^{-1/5}}{2^4(1-\frac{3}{2}x)^4}$ $E = \frac{1}{16} (1+\frac{2}{3}x)^{-3} (1-15x)^{-1/5} (1-\frac{3}{2}x)^{-4}$ Applying the approximation $(1+ax)^n \approx 1+nax$: $(1+\frac{2}{3}x)^{-3} \approx 1 + (-3)(\frac{2}{3}x) = 1-2x$ $(1-15x)^{-1/5} \approx 1 + (-\frac{1}{5})(-15x) = 1+3x$ $(1-\frac{3}{2}x)^{-4} \approx 1 + (-4)(-\frac{3}{2}x) = 1+6x$ Multiplying these: $E \approx \frac{1}{16} (1-2x)(1+3x)(1+6x)$ $E \approx \frac{1}{16} (1+3x-2x)(1+6x) = \frac{1}{16} (1+x)(1+6x)$ $E \approx \frac{1}{16} (1+6x+x) = \frac{1}{16}(1+7x)$

If $x$ is so small that $x^2$ and higher powers of $x$ may be neglected,then the approximate value of $\frac{(1+\frac{2}{3}x)^{-3}(1-15x)^{-1/5}}{(2-3x)^4}$ is:

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