Страница 117 номер 405, ГДЗ по алгебре за 9 класс к учебнику Мерзляка
1) \(x^{2} + 4x + 3 > 0\)
\(a > 0\) ветви вверх
\(\dfrac{D}{4} = 4 - 3 = 1\)
\(x_{1} = - 2 - 1 = - 3;\ x_{2} = - 2 + 1 = - 1\)
\(x \in ( - \infty; - 3) \cup ( - 1; + \infty)\)
2) \(x^{2} - 3x + 2 \leqslant 0\)
\(a > 0\) ветви вверх
\(D = 9 - 4 \cdot 2 = 1\)
\(x_{1} = \dfrac{3 - 1}{2} = 1;\ x_{2} = \dfrac{3 + 1}{2} = 2\)
\(x \in [ 1;\ 2]\)
3) \(-x^{2} + 12x + 45 < 0\)
\(a < 0\) ветви вниз
\(\dfrac{D}{4} = 36 + 45 = 81 = 9^{2}\)
\(x_{1} = \dfrac{- 6 - 9}{- 1} = 15;\ x_{2} = \dfrac{- 6 + 9}{- 1} = - 3\)
\(x \in ( - \infty; - 3) \cup (15; + \infty)\)
4) \(- 3x^{2} - 5x - 2 \geqslant 0\)
\(a < 0\) ветви вниз
\(D = 25 + 4 \cdot ( - 3) \cdot 2 = 25 - 24 = 1\)
\(x_{1} = \dfrac{5 - 1}{- 6} = - \dfrac{4}{6} = - \dfrac{2}{3};\ x_{2} = \dfrac{5 + 1}{- 6} = - 1\)
\(x \in \left[ - 1; - \dfrac{2}{3} \right]\)
5) \(x^{2} - 5x > 0\)
\(a > 0\) ветви вверх
\(x(x - 5) > 0 \Rightarrow x_{1} = 0;\ x_{2} = 5 \Rightarrow D > 0\)
\(x \in ( - \infty;\ 0) \cup (5; + \infty)\)
6) \(- 25x^{2} + 16 \leqslant 0\)
\(a < 0\) ветви вниз
\(D = 0 - 4 \cdot ( - 25) \cdot 16 = 2^{2} \cdot 5^{2} \cdot 4^{2} = 40^{2}\)
\(x_{1} = \dfrac{- 40}{- 50} = \dfrac{4}{5};\ x_{2} = \dfrac{40}{- 50} = - \dfrac{4}{5}\)
\(x \in \left( - \infty; - \dfrac{4}{5} \right] \cup \left[ \dfrac{4}{5}; + \infty \right)\)
7) \(5x^{2} - 3x + 1 \geqslant 0\)
\(a > 0\) ветви вверх
\(D = 9 - 4 \cdot 5 < 0\) нет корней
\(x \in ( - \infty; + \infty)\)
8) \(- 3x^{2} + 6x - 4 > 0\)
\(a < 0\) ветви вниз
\(D = 36 + 4 \cdot ( - 3) \cdot 4 < 0\) нет корней
\(x \in \varnothing\)
9) \(\frac{1}{3}x^{2} - 2x + 3 \leqslant 0\)
\(a > 0\) ветви вверх
\(D = 4 - 4 \cdot \dfrac{1}{3} \cdot 3 = 0\)
\(x = \dfrac{2}{2 \cdot \dfrac{1}{3}} = 3\)
\(x = 3\)
10) \(-x^{2} + \frac{1}{3}x - \frac{1}{36} > 0\)
\(a < 0\) ветви вниз
\(D = \dfrac{1}{9} + 4 \cdot \dfrac{1}{36} \cdot ( - 1) = \dfrac{1}{9} - \dfrac{1}{9} = 0\)
\(x = \dfrac{- 1/3}{- 2} = \dfrac{1}{6}\)
\(x \in \varnothing\)
11) \(2x^{2} - 2x + 0{,}5 < 0\)
\(a > 0\) ветви вверх
\(\dfrac{D}{4} = 1 - 2 \cdot 0{,}5 = 0\)
\(x = \dfrac{1}{2} = 0{,}5\)
\(x \in \varnothing\)
Ответ: 1) \(x \in ( - \infty; - 3) \cup ( - 1; + \infty)\); 2) \(x \in [ 1;\ 2]\); 3) \(x \in ( - \infty; - 3) \cup (15; + \infty)\); 4) \(x \in \left[ - 1; - \frac{2}{3} \right]\); 5) \(x \in ( - \infty;\ 0) \cup (5; + \infty)\); 6) \(x \in \left( - \infty; - \frac{4}{5} \right] \cup \left[ \frac{4}{5}; + \infty \right)\); 7) \(x \in ( - \infty; + \infty)\); 8) \(x \in \varnothing\); 9) \(x = 3\); 10) \(x \in \varnothing\); 11) \(x \in \varnothing\).
