Решите неравенство 2/(x^2+10x+27)+5/(x^2+10x+26)≥6.

Ответ:

\[\frac{2}{x^{2} + 10x + 27} + \frac{5}{x^{2} + 10x + 26} \geq 6\]

\[y = x^{2} + 10x + 26 = (x + 5)^{2} + 1 \geq 1:\]

\[\frac{2}{y + 1} + \frac{5}{y} \geq 6\ \ \ \ \ \ | \cdot y(y + 1)\]

\[2y + 5 \cdot (y + 1) \geq 6 \cdot \left( y^{2} + y \right)\]

\[2y + 5y + 5 \geq 6y^{2} + 6y\]

\[6y^{2} - y - 5 \leq 0\]

\[D = 1 + 120 = 121\]

\[y_{1} = \frac{1 + 11}{12} = 1;\ \ \]

\[y_{2} = \frac{1 - 11}{12} = - \frac{10}{12} = - \frac{5}{6}\]

\[6\left( y + \frac{5}{6} \right)(y - 1) \leq 0\]

\[y \in \left\lbrack - \frac{5}{6};1 \right\rbrack.\]

\[Так\ как\ y \geq 1;то\ получаем:y = 1.\]

\[Подставим:\]

\[(x + 5)^{2} + 1 = 1\]

\[(x + 5)^{2} = 0\]

\[x + 5 = 0\]

\[x = - 5.\]

\[Ответ:x = - 5.\]