Решебник по алгебре 9 класс Макарычев Задание 376

 

\[\boxed{\text{376\ (376).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ x^{2} - 5x - 50 < 0\]

\[По\ теореме\ Виета:\]

\[x_{1} + x_{2} = 5;\ \ \ \ x_{1} \cdot x_{2} = - 50\]

\[x_{1} = 10;\ \ x_{2} = - 5.\]

\[(x - 10)(x + 5) < 0\]

\[x \in ( - 5;10).\]

\[\textbf{б)} - m^{2} - 8x + 9 \geq 0\]

\[m^{2} + 8m - 9 \leq 0\]

\[D_{1} = 16 + 9 = 25\]

\[m_{1} = - 4 - 5 = - 9;\ \ \ \]

\[m_{2} = - 4 + 5 = 1.\]

\[(m + 9)(m - 1) \leq 0\]

\[m \in \lbrack - 9;1\rbrack.\]

\[\textbf{в)}\ 3y^{2} + 4y - 4 > 0\]

\[D_{1} = 4 + 3 \cdot 4 = 16\]

\[y_{1,2} = \frac{- 2 \pm 4}{3} = - 2;\ \frac{2}{3}.\]

\[3 \cdot (x + 2)\left( x - \frac{2}{3} \right) > 0\ \]

\[y \in ( - \infty;\ - 2) \cup \left( \frac{2}{3};\ + \infty \right).\]

\[\textbf{г)}\ 8p^{2} + 2p \geq 21\]

\[8p^{2} + 2p - 21 \geq 0\]

\[D_{1} = 1 + 8 \cdot 21 = 169\]

\[p_{1,2} = \frac{- 1 \pm 13}{8} = 1,5;\ - 1,75.\]

\[8 \cdot (x + 1,75)(x - 1,5) \geq 0\]

\[p \in ( - \infty; - 1,75\rbrack \cup \lbrack 1,5;\ + \infty).\]

\[\textbf{д)}\ 12x - 9 \leq 4x^{2}\]

\[4x^{2} - 12x + 9 \geq 0\]

\[(2x - 3)^{2} \geq 0\]

\[x \in ( - \infty; + \infty).\]

\[\textbf{е)} - 9x^{2} < 1 - 6x\]

\[9x^{2} - 6x + 1 > 0\ \]

\[(3x - 1)^{2} > 0\]

\[3 \cdot \left( x - \frac{1}{3} \right)^{2} > 0\]

\[x \in \left( - \infty;\frac{1}{3} \right) \cup \left( \frac{1}{3}; + \infty \right).\]

\[\boxed{\text{376.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[B(a;1 - a)\]

\[\textbf{а)}\ x^{2} - y^{2} = 14\]

\[a^{2} - (1 - a)^{2} = 14\]

\[a^{2} - 1 + 2a - a^{2} = 14\]

\[2a = 15\]

\[a = 7,5.\]

\[При\ a = 7,5.\]

\[\textbf{б)}\ x^{2} + y^{2} = 1\]

\[a^{2} + (1 - a)^{2} = 1\]

\[a^{2} + 1 - 2a + a^{2} = 1\]

\[2a^{2} - 2a = 0\]

\[2a(a - 1) = 0\]

\[a = 0;\ \ a = 1.\]

\[При\ \ a = 0;\ \ a = 1.\ \]