Question

See attachment the problem can be found in there

Answer 1

Answer:

$\frac{ - 2 {x}^{2} + 19x + 3 }{3 {x} (4x^{2} - 9 )}$

Step-by-step explanation:

$\frac{5}{6 {x}^{2} + 9x} + \frac{1}{2x - 3} - \frac{2}{3x} \\ \\ = \frac{5}{3x(2 {x} + 3)} + \frac{1}{2x - 3} - \frac{2}{3x} \\ \\ = \frac{5(2x - 3) + 1 \times 3x(2x + 3)}{3x(2 {x} + 3)(2x - 3)} - \frac{2}{3x} \\ \\ = \frac{10x - 15 + 6 {x}^{2} + 9x}{3x((2 {x} ) ^{2} - {3}^{2} )} - \frac{2}{3x} \\ \\ = \frac{6 {x}^{2} + 19x - 15}{3x(4x^{2} - 9 )} - \frac{2}{3x} \\ \\ = \frac{3x(6 {x}^{2} + 19x - 15) - 2 \times 3x(4x^{2} - 9 )}{3x(4x^{2} - 9 ) \times 3x} \\ \\ = \frac{18{x}^{3} + 57 {x}^{2} - 45x - 24x^{3} + 54x }{3x(4x^{2} - 9 ) \times 3x} \\ \\ = \frac{ - 6 {x}^{3} + 57 {x}^{2} + 9x }{9 {x}^{2} (4x^{2} - 9 )} \\ \\ = \frac{ 3x(- 2 {x}^{2} + 19x + 3) }{9 {x}^{2} (4x^{2} - 9 )} \\ \\ \huge \orange{ \boxed{= \frac{ - 2 {x}^{2} + 19x + 3 }{3 {x} (4x^{2} - 9 )} }}$