Question

$18x^{2} - 11x + 1 > 0$

$$(2x - 1)(9x - 1) > 0 $$$x < \frac{1}{9}$ or $\frac{1}{2} < x$

Also $10x^{2} - 17x + 7 > 0$

$$(x - 1)(10x - 7) > 0 $$$x < \frac{7}{10}$ or $1 < x$

$${\& 10x^{2} - 17x + 7 \neq 1 }{x \in \left( - \infty,\frac{1}{9} \right) \cup \left( \frac{1}{2},\frac{7}{10} \right) \cup (1,\infty) - \left\{ \frac{6}{5} \right\} }{90(a + b + c + d + e) = 90\left( \frac{1}{9} + \frac{1}{2} + \frac{7}{10} + 1 + \frac{6}{5} \right) }{= 10 + 45 + 63 + 90 + 108 = 316}$$