Решите неравенство: x^3-5x^2+6x>=0.

Ответ:

\[x^{3} - 5x^{2} + 6x \geq 0\]

\[x\left( x^{2} - 5x + 6 \right) \geq 0\]

\[x^{2} - 5x + 6 = 0\]

\[x_{1} + x_{2} = 5;\ \ \ x_{1} \cdot x_{2} = 6\]

\[x_{1} = 2;\ \ \ x_{2} = 3\]

\[x(x - 2)(x - 3) \geq 0\]

\[0 \leq x \leq 2;\ \ x \geq 3.\]