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(D) Given expression is $f(x) = \frac{(3-5x)^{1/2}}{(5-3x)^2} = (3-5x)^{1/2} \cdot (5-3x)^{-2}$.We can rewrite this as $f(x) = \sqrt{3}(1-\frac{5x}{3}

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$\frac{1+30\sqrt{3}}{750}$

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(D) Given expression is $f(x) = \frac{(3-5x)^{1/2}}{(5-3x)^2} = (3-5x)^{1/2} \cdot (5-3x)^{-2}$.We can rewrite this as $f(x) = \sqrt{3}(1-\frac{5x}{3})^{1/2} \cdot \frac{1}{25}(1-\frac{3x}{5})^{-2} = \frac{\sqrt{3}}{25}(1-\frac{5x}{3})^{1/2}(1-\frac{3x}{5})^{-2}$.Using the binomial approximation $(1+u)^n \approx 1+nu$ for small $|u|$,we have:$(1-\frac{5x}{3})^{1/2} \approx 1 + \frac{1}{2}(-\frac{5x}{3}) = 1 - \frac{5x}{6}$.$(1-\frac{3x}{5})^{-2} \approx 1 + (-2)(-\frac{3x}{5}) = 1 + \frac{6x}{5}$.Multiplying these approximations and neglecting $x^2$ terms:$f(x) \approx \frac{\sqrt{3}}{25}(1 - \frac{5x}{6})(1 + \frac{6x}{5}) \approx \frac{\sqrt{3}}{25}(1 + \frac{6x}{5} - \frac{5x}{6}) = \frac{\sqrt{3}}{25}(1 + \frac{36x-25x}{30}) = \frac{\sqrt{3}}{25}(1 + \frac{11x}{30})$.Given $x = \frac{1}{\sqrt{363}} = \frac{1}{11\sqrt{3}}$.Substituting $x$ into the expression:$f(x) \approx \frac{\sqrt{3}}{25}(1 + \frac{11}{30} \cdot \frac{1}{11\sqrt{3}}) = \frac{\sqrt{3}}{25}(1 + \frac{1}{30\sqrt{3}}) = \frac{\sqrt{3}}{25} + \frac{\sqrt{3}}{25 \cdot 30\sqrt{3}} = \frac{\sqrt{3}}{25} + \frac{1}{750} = \frac{30\sqrt{3}+1}{750}$.

If $|x|$ is so small that all terms containing $x^2$ and higher powers of $x$ can be neglected,then the approximate value of $\frac{(3-5x)^{1/2}}{(5-3x)^2}$,when $x=\frac{1}{\sqrt{363}}$,is

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