Для любого числа х € R докажите справедливость неравенства: (x^2-4x-5)/5+5/(x^2-4x+5)>=0 найдите значения x, при которых левая часть неравенства равна правой.

\[\frac{x^{2} - 4x - 5}{5} + \frac{5}{x^{2} - 4x + 5} \geq 0\]

\[Пусть\ t = x^{2} - 4x - 5:\]

\[\frac{t}{5} + \frac{5}{t + 10} \geq 0\]

\[\frac{t(t + 10) + 25}{5 \cdot (t + 10)} \geq 0\]

\[\frac{t^{2} + 10t + 25}{5 \cdot (t + 10)} \geq 0\]

\[\frac{(t + 5)^{2}}{5 \cdot (t + 10)} \geq 0\]

\[\frac{\left( x^{2} - 4x - 5 + 5 \right)^{2}}{5 \cdot \left( x^{2} - 4x - 5 + 10 \right)} \geq 0\]

\[\frac{\left( x^{2} - 4x \right)^{2}}{5 \cdot \left( x^{2} - 4x + 5 \right)} \geq 0\]

\[x^{2} - 4x = 0\]

\[x(x - 4) = 0\]

\[x_{1} = 0;\ \ \ \ \ \ x_{2} = 4.\]

\[Ответ:\ \ x = 0;x = 4.\]