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МатематикаРешебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 39
Задание 39
\[\boxed{\mathbf{39.}}\]
\[\textbf{а)}\ \sqrt{x + 3}\sqrt{x + 8} > 2x + 4\]
\[x + 3 \geq 0\]
\[x \geq - 3.\]
\[x + 8 \geq 0\]
\[x \geq - 8.\]
\[M = \lbrack - 3; + \infty).\]
\[(x + 3)(x + 8) > (2x + 4)^{2}\]
\[x^{2} + 3x + 8x + 24 >\]
\[> 4x^{2} + 16x + 16\]
\[3x^{2} + 5x - 8 < 0\]
\[D = 25 + 96 = 121\]
\[x_{1} = \frac{- 5 + 11}{6} = 1;\]
\[x_{2} = \frac{- 5 - 11}{6} = - \frac{16}{6} = - \frac{8}{3} =\]
\[= - 2\frac{2}{3}.\]
\[\left( x + 2\frac{2}{3} \right)(x - 1) < 0\]
\[- 2\frac{2}{3} < x < 1.\]
\[Решение\ неравенства:\]
\[x \in \lbrack - 3;1).\]
\[Ответ:x \in \lbrack - 3;1).\]
\[\textbf{б)}\ \sqrt{x + 3}\sqrt{x + 6} < x + 4\]
\[x + 3 \geq 0\]
\[x \geq - 3.\]
\[x + 6 \geq 0\]
\[x \geq - 6.\]
\[M = \lbrack - 3;\ + \infty).\]
\[(x + 3)(x + 6) < (x + 4)^{2}\]
\[x^{2} + 3x + 6x + 18 <\]
\[< x^{2} + 8x + 16\]
\[x < - 2.\]
\[Решение\ неравенства:\]
\[x \in \lbrack - 3; - 2)\]
\[Ответ:x \in \lbrack - 3; - 2).\]
\[\textbf{в)}\ \sqrt{2x + 3}\sqrt{3x + 7} > 2x + 4\]
\[2x + 3 \geq 0\]
\[2x \geq - 3\]
\[x \geq - 1,5.\]
\[3x + 7 \geq 0\]
\[3x \geq - 7\]
\[x \geq - 2\frac{1}{3}.\]
\[M = \lbrack - 1,5;\ + \infty).\]
\[(2x + 3)(3x + 7) > (2x + 4)^{2}\]
\[6x^{2} + 9x + 14x + 21 >\]
\[> 4x^{2} + 16x + 16\]
\[2x^{2} + 7x + 5 > 0\]
\[D = 49 - 40 = 9\]
\[x_{1} = \frac{- 7 + 3}{4} = - 1;\]
\[x_{2} = \frac{- 7 - 3}{4} = - 2,5.\]
\[(x + 2,5)(x + 1) > 0\]
\[x < - 2,5;\ \ x > - 1.\]
\[Решение\ неравенства:\]
\[x \in ( - 1; + \infty).\]
\[Ответ:x \in ( - 1; + \infty).\]
\[\textbf{г)}\ \sqrt{2x - 1}\sqrt{3x - 2} < 4x - 3\]
\[2x - 1 \geq 0\]
\[2x \geq 1\]
\[x \geq 0,5.\]
\[3x - 2 \geq 0\]
\[3x \geq 2\]
\[x \geq \frac{2}{3}.\]
\[M = \left\lbrack \frac{2}{3}; + \infty \right).\]
\[(2x - 1)(3x - 2) < (4x - 3)^{2}\]
\[6x^{2} - 3x - 4x + 2 <\]
\[< 16x^{2} - 24x + 9\]
\[10x^{2} - 17x + 7 > 0\]
\[D = 289 - 280 = 9\]
\[x_{1} = \frac{17 + 3}{20} = 1;\]
\[x_{2} = \frac{17 - 3}{20} = \frac{7}{10} = 0,7;\]
\[(x - 0,7)(x - 1) > 0\]
\[x < 0,7;\ \ x > 1.\]
\[Решение\ неравенства:\]
\[x \in (1; + \infty).\]
\[Ответ:x \in (1; + \infty).\]
