5) \(\frac{5}{x^2+3x} - \frac{15}{x^2-3x} = \frac{16}{x}\);

5) \(\frac{5}{x^2+3x} - \frac{15}{x^2-3x} = \frac{16}{x}\);

ОДЗ: \(x
eq 0\), \(x
eq \pm 3\)

\(\frac{5}{x(x+3)} - \frac{15}{x(x-3)} = \frac{16}{x}\)

\(\frac{5(x-3) - 15(x+3)}{x(x+3)(x-3)} = \frac{16}{x}\)

\(\frac{5x - 15 - 15x - 45}{x(x^2 - 9)} = \frac{16}{x}\)

\(\frac{-10x - 60}{x(x^2-9)} = \frac{16}{x}\)

\(-10x - 60 = 16(x^2 - 9)\)

\(-10x - 60 = 16x^2 - 144\)

\(16x^2 + 10x - 84 = 0\)

\(8x^2 + 5x - 42 = 0\)

\(D = 5^2 - 4 \cdot 8 \cdot (-42) = 25 + 1344 = 1369 = 37^2\)

\(x_1 = \frac{-5 + 37}{2 \cdot 8} = \frac{32}{16} = 2\)

\(x_2 = \frac{-5 - 37}{2 \cdot 8} = \frac{-42}{16} = -\frac{21}{8}\)

Ответ: -21/8; 2