Решите неравенство 3/(x^2+8x+17)+4/(x^2+8x+18)≥5.

Ответ:

\[\frac{3}{x^{2} + 8x + 17} + \frac{4}{x^{2} + 8x + 18} \geq 5\]

\[y = x^{2} + 8x + 17 = (x + 4)^{2} + 1 \geq 1.\]

\[\frac{3}{y} + \frac{4}{y + 1} \geq 5\ \ \ \ \ \ \ \ | \cdot y(y + 1)\]

\[3(y + 1) + 4y \geq 5\left( y^{2} + y \right)\]

\[3y + 3 + 4y \geq 5y^{2} + 5y\]

\[5y^{2} - 2y - 3 \leq 0\]

\[D = 4 + 60 = 64\]

\[y_{1} = \frac{2 + 8}{10} = 1;\ \ y_{2} = \frac{2 - 8}{10} = - 0,6\]

\[5(y + 0,6)(y - 1) \leq 0\]

\[y \in \lbrack - 0,6;1\rbrack\text{.\ }\]

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

\[Подставим:\]

\[(x + 4)^{2} + 1 = 1\]

\[(x + 4)^{2} = 0\]

\[x + 4 = 0\]

\[x = - 4.\]

\[Ответ:x = - 4.\]