Страница 117 номер 408, ГДЗ по алгебре за 9 класс к учебнику Мерзляка
1) \(x(x + 5) - 2 < 4x\)
\(x^{2} + 5x - 2 - 4x < 0\)
\(x^{2} + x - 2 < 0\)
\((x - 1)(x + 2) < 0\)
\(x \in ( - 2;\ 1)\)
2) \(11 - (x + 1)^{2} \leqslant x\)
\(11 - x^{2} - 2x - 1 - x \leqslant 0\)
\(- x^{2} - 3x + 10 \leqslant 0\)
\(D = 9 - 4 \cdot ( - 1) \cdot 10 = 49 = 7^{2}\)
\(x_{1} = \dfrac{3 - 7}{- 2} = 2;\ x_{2} = \dfrac{3 + 7}{- 2} = - 5\)
\(x \in ( - \infty; - 5] \cup [ 2; + \infty)\)
3) \((2x + 1)^{2} - (x + 1)(x - 7) \leqslant 5\)
\(4x^{2} + 4x + 1 - x^{2} + 6x + 7 - 5 \leqslant 0\)
\(3x^{2} + 10x + 3 \leqslant 0\)
\(\dfrac{D}{4} = 25 - 9 = 16 = 4^{2}\)
\(x_{1} = \dfrac{- 5 - 4}{3} = - 3;\ x_{2} = \dfrac{- 5 + 4}{3} = - \dfrac{1}{3}\)
\(x \in \left[ - 3; - \dfrac{1}{3} \right]\)
4) \(5x(x + 4) - (2x - 3)(2x + 3) > 30\)
\(5x^{2} + 20x - 4x^{2} + 9 - 30 > 0\)
\(x^{2} + 20x - 21 > 0\)
\((x - 1)(x + 21) > 0\)
\(x \in ( - \infty; - 21) \cup (1; + \infty)\)
5) \((3x - 7)(x + 2) - (x - 4)(x + 5) > 30\)
\(3x^{2} - 7x + 6x - 14 - x^{2} - x + 20 - 30 > 0\)
\(2x^{2} - 2x - 24 > 0\)
\(x^{2} - x - 12 > 0\)
\((x + 3)(x - 4) > 0\)
\(x \in ( - \infty; - 3) \cup (4; + \infty)\)
6) \(\frac{2x^{2} - 1}{4} - \frac{3 - 4x}{6} + \frac{8x - 5}{8} \leqslant \frac{19}{24}\)
\(6\left( 2x^{2} - 1 \right) - 4(3 - 4x) + 3(8x - 5) \leqslant 19\)
\(12x^{2} - 6 - 12 + 16x + 24x - 15 - 19 \leqslant 0\)
\(12x^{2} + 40x - 52 \leqslant 0\)
\(3x^{2} + 10x - 13 \leqslant 0\)
\(\dfrac{D}{4} = 25 + 3 \cdot 13 = 25 + 39 = 64 = 8^{2}\)
\(x_{1} = \dfrac{- 5 - 8}{3} = - \dfrac{13}{3};\ x_{2} = \dfrac{- 5 + 8}{3} = 1\)
\(x \in \left[ - \dfrac{13}{3};\ 1 \right]\)
Ответ: 1) \(x \in ( - 2;\ 1)\); 2) \(x \in ( - \infty; - 5] \cup [ 2; + \infty)\); 3) \(x \in \left[ - 3; - \frac{1}{3} \right]\); 4) \(x \in ( - \infty; - 21) \cup (1; + \infty)\); 5) \(x \in ( - \infty; - 3) \cup (4; + \infty)\); 6) \(x \in \left[ - \frac{13}{3};\ 1 \right]\).
