26 февраля 2017
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Задача 15
Решите неравенство
\(\frac{1}{{\sqrt { - x - 2} }} - \frac{1}{{\sqrt {x + 4} }} \le 1 + \frac{1}{{\sqrt {\left( {x + 4} \right)\left( {x - 2} \right)} }}.\)
Решение
\(\frac{1}{{\sqrt { - x - 2} }} - \frac{1}{{\sqrt {x + 4} }} \le 1 + \frac{1}{{\sqrt {\left( {x + 4} \right)\left( {x - 2} \right)} }} \Leftrightarrow \)
\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{\sqrt {x + 4} - \sqrt { - x - 2} \le 1 + \sqrt {\left( {x + 4} \right)\left( {x - 2} \right)} ,}\\{ - x - 2 > 0,}\\{x + 4 > 0.}\end{array}} \right.\)
Рассмотрим два случая:
1) \(\left\{ {\begin{array}{*{20}{c}}{\sqrt {x + 4} < \sqrt { - x - 2} ,}\\{ - 4 < x < - 2;}\end{array}} \right. \Leftrightarrow - 4 < x < - 3;\)
2) \(\left\{ {\begin{array}{*{20}{c}}{\sqrt {x + 4} \ge \sqrt { - x - 2} ,}\\{{{\left( {\sqrt {x + 4} - \sqrt { - x - 2} } \right)}^2} \le {{\left( {1 + \sqrt {\left( {x + 4} \right)\left( {x - 2} \right)} } \right)}^2},}\\{ - 4 < x < - 2;}\end{array}} \right. \Leftrightarrow \)
\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{ - 3 \le x < - 2,}\\{{t^2} + 4t - 1 \ge 0,\;\;t = \sqrt {\left( {x + 4} \right)\left( {x - 2} \right)} > 0;}\end{array}} \right. \Leftrightarrow \)
\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{ - 3 \le x < - 2,}\\{\left( {t - {t_1}} \right)\left( {t - {t_2}} \right) \ge 0,\;\;{t_{1,2}} = - 2 \pm \sqrt 5 ;}\end{array}} \right. \Leftrightarrow \)
\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{ - 3 \le x < - 2,}\\{\sqrt {\left( {x + 4} \right)\left( {x - 2} \right)} \ge - 2 + \sqrt 5 ;}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{ - 3 \le x < - 2,}\\{{x^2} + 6x + 17 - 4\sqrt 5 \le 0;}\end{array}} \right. \Leftrightarrow \)
\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{ - 3 \le x < - 2,}\\{\left( {x - {x_1}} \right)\left( {x - {x_2}} \right) \le 0,\;\;{x_{1,2}} = - 3 \pm 2\sqrt {\sqrt 5 - 2} ;}\end{array}} \right. \Leftrightarrow \)
\( \Leftrightarrow - 3 \le x \le - 3 + 2\sqrt {\sqrt 5 - 2} ,\)
так как \( - 3 + 2\sqrt {\sqrt 5 - 2} < - 2 \Leftrightarrow 2\sqrt {\sqrt 5 - 2} < 1 \Leftrightarrow 4\sqrt 5 - 8 < - 1 \Leftrightarrow \)
\( \Leftrightarrow 80 < 81,\)
что верно.
Ответ: \( - 4 < x \le - 3 + 2\sqrt {\sqrt 5 - 2} \).
\(\frac{1}{{\sqrt { - x - 2} }} - \frac{1}{{\sqrt {x + 4} }} \le 1 + \frac{1}{{\sqrt {\left( {x + 4} \right)\left( {x - 2} \right)} }}.\)
Решение
\(\frac{1}{{\sqrt { - x - 2} }} - \frac{1}{{\sqrt {x + 4} }} \le 1 + \frac{1}{{\sqrt {\left( {x + 4} \right)\left( {x - 2} \right)} }} \Leftrightarrow \)
\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{\sqrt {x + 4} - \sqrt { - x - 2} \le 1 + \sqrt {\left( {x + 4} \right)\left( {x - 2} \right)} ,}\\{ - x - 2 > 0,}\\{x + 4 > 0.}\end{array}} \right.\)
Рассмотрим два случая:
1) \(\left\{ {\begin{array}{*{20}{c}}{\sqrt {x + 4} < \sqrt { - x - 2} ,}\\{ - 4 < x < - 2;}\end{array}} \right. \Leftrightarrow - 4 < x < - 3;\)
2) \(\left\{ {\begin{array}{*{20}{c}}{\sqrt {x + 4} \ge \sqrt { - x - 2} ,}\\{{{\left( {\sqrt {x + 4} - \sqrt { - x - 2} } \right)}^2} \le {{\left( {1 + \sqrt {\left( {x + 4} \right)\left( {x - 2} \right)} } \right)}^2},}\\{ - 4 < x < - 2;}\end{array}} \right. \Leftrightarrow \)
\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{ - 3 \le x < - 2,}\\{{t^2} + 4t - 1 \ge 0,\;\;t = \sqrt {\left( {x + 4} \right)\left( {x - 2} \right)} > 0;}\end{array}} \right. \Leftrightarrow \)
\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{ - 3 \le x < - 2,}\\{\left( {t - {t_1}} \right)\left( {t - {t_2}} \right) \ge 0,\;\;{t_{1,2}} = - 2 \pm \sqrt 5 ;}\end{array}} \right. \Leftrightarrow \)
\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{ - 3 \le x < - 2,}\\{\sqrt {\left( {x + 4} \right)\left( {x - 2} \right)} \ge - 2 + \sqrt 5 ;}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{ - 3 \le x < - 2,}\\{{x^2} + 6x + 17 - 4\sqrt 5 \le 0;}\end{array}} \right. \Leftrightarrow \)
\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{ - 3 \le x < - 2,}\\{\left( {x - {x_1}} \right)\left( {x - {x_2}} \right) \le 0,\;\;{x_{1,2}} = - 3 \pm 2\sqrt {\sqrt 5 - 2} ;}\end{array}} \right. \Leftrightarrow \)
\( \Leftrightarrow - 3 \le x \le - 3 + 2\sqrt {\sqrt 5 - 2} ,\)
так как \( - 3 + 2\sqrt {\sqrt 5 - 2} < - 2 \Leftrightarrow 2\sqrt {\sqrt 5 - 2} < 1 \Leftrightarrow 4\sqrt 5 - 8 < - 1 \Leftrightarrow \)
\( \Leftrightarrow 80 < 81,\)
что верно.
Ответ: \( - 4 < x \le - 3 + 2\sqrt {\sqrt 5 - 2} \).