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(D) We have, $\frac{5x^{2}-26x+5}{3x^{2}-10x+3} < 0$Factorizing the numerator and denominator:$\frac{5x^{2}-25x-x+5}{3x^{2}-9x-x+3} < 0$$\frac{5x(x-5)

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$\frac{1}{5} < x < \frac{1}{3} \cup 3 < x < 5$

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(D) We have, $\frac{5x^{2}-26x+5}{3x^{2}-10x+3} Factorizing the numerator and denominator:$\frac{5x^{2}-25x-x+5}{3x^{2}-9x-x+3} $\frac{5x(x-5)-1(x-5)}{3x(x-3)-1(x-3)} $\frac{(x-5)(5x-1)}{(x-3)(3x-1)} The critical points are $x = \frac{1}{5}, \frac{1}{3}, 3, 5$.Using the wavy curve method (sign scheme) for the expression $f(x) = \frac{(x-5)(5x-1)}{(x-3)(3x-1)}$, the expression is negative in the intervals $(\frac{1}{5}, \frac{1}{3})$ and $(3, 5)$.Therefore, $x \in (\frac{1}{5}, \frac{1}{3}) \cup (3, 5)$.

Given that $x$ is a real number satisfying $\frac{5x^{2}-26x+5}{3x^{2}-10x+3} < 0$, then

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