\[ x^2 - 9 = (x - 3)(x + 3) \]
\[ \frac{1 \cdot (x+3)}{(x-3)(x+3)} + \frac{18}{(x-3)(x+3)} = \frac{x \cdot (x-3)}{(x-3)(x+3)} \]
\[ (x+3) + 18 = x(x-3) \]
\[ x + 3 + 18 = x^2 - 3x \]
\[ x + 21 = x^2 - 3x \]
\[ x^2 - 3x - x - 21 = 0 \]
\[ x^2 - 4x - 21 = 0 \]
\[ D = b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot (-21) = 16 + 84 = 100 \]
\[ \sqrt{D} = 10 \]
\[ x_1 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - 10}{2 \cdot 1} = \frac{-6}{2} = -3 \]
\[ x_2 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + 10}{2 \cdot 1} = \frac{14}{2} = 7 \]
При x = -3:
x - 3 = -3 - 3 = -6
eq 0
x + 3 = -3 + 3 = 0
x^2 - 9 = (-3)^2 - 9 = 9 - 9 = 0
Так как при x = -3 знаменатели обращаются в ноль, этот корень не подходит.
При x = 7:
x - 3 = 7 - 3 = 4
eq 0
x + 3 = 7 + 3 = 10
eq 0
x^2 - 9 = 7^2 - 9 = 49 - 9 = 40
eq 0
Ответ: 7