Решите уравнение: (корень из (x)-3)(18x^2-9x-5)=0.

Ответ:

\[\left( \sqrt{x} - 3 \right)\left( 18x^{2} - 9x - 5 \right) = 0\]

\[\sqrt{x} = 3 \rightarrow \ \ x = 9.\]

\[D = ( - 9)^{2} - 4 \cdot 18 \cdot ( - 5) =\]

\[= 81 + 360 = 441\]

\[x_{1} = \frac{9 + \sqrt{441}}{2 \cdot 18} = \frac{9 + 21}{36} =\]

\[= \frac{30}{36} = \frac{5}{6}\]

\[x_{2} = \frac{9 - \sqrt{441}}{2 \cdot 18} = \frac{9 - 21}{36} =\]

\[= - \frac{12}{36} = - \frac{1}{3}\]

\[Ответ:x = 9;x = \frac{5}{6};\ x = - \frac{1}{3}.\]