Страница 117 номер 409, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

1) \(2\left( x^{2} + 2 \right) \geqslant x(x + 5)\)
\(2x^{2} + 4 \geqslant x^{2} + 5x\)
\(x^{2} - 5x + 4 \geqslant 0\)
\((x - 1)(x - 4) \geqslant 0\)

\(x \in ( - \infty;\ 1] \cup [ 4; + \infty)\)

2) \(x - (x + 4)(x + 5) > - 5\)
\(x - x^{2} - 9x - 20 + 5 > 0\)
\(- x^{2} - 8x - 15 > 0\)
\(\dfrac{D}{4} = 16 - 15 = 1\)
\(x_{1} = \dfrac{4 - 1}{- 1} = - 3;\ x_{2} = \dfrac{4 + 1}{- 1} = - 5\)

\(x \in ( - 5; - 3)\)

3) \((6x - 1)(6x + 1) - (12x - 5)(x + 2) < 7 - 3x\)
\(36x^{2} - 1 - 12x^{2} + 5x - 24x + 10 - 7 + 3x < 0\)
\(24x^{2} - 16x + 2 < 0\)
\(12x^{2} - 8x + 1 < 0\)
\(\dfrac{D}{4} = 16 - 12 = 4\)
\(x_{1} = \dfrac{4 - 2}{12} = \dfrac{2}{12} = \dfrac{1}{6};\ x_{2} = \dfrac{4 + 2}{12} = \dfrac{6}{12} = \dfrac{1}{2}\)

\(x \in \left( \dfrac{1}{6};\dfrac{1}{2} \right)\)

4) \(\frac{x - 1}{4} - \frac{2x - 3}{2} < \frac{x^{2} + 3x}{8}\)
\(2(x - 1) - 4(2x - 3) < x^{2} + 3x\)
\(2x - 2 - 8x + 12 - x^{2} - 3x < 0\)
\(- x^{2} - 9x + 10 < 0\)
\(- \left( x^{2} + 9x - 10 \right) < 0\)
\(- (x - 1)(x + 10) < 0\)

\(x \in ( - \infty; - 10) \cup (1; + \infty)\)

Ответ: 1) \(x \in ( - \infty;\ 1] \cup [ 4; + \infty)\); 2) \(x \in ( - 5; - 3)\); 3) \(x \in \left( \frac{1}{6};\frac{1}{2} \right)\); 4) \(x \in ( - \infty; - 10) \cup (1; + \infty)\).