Решебник по алгебре 8 класс Мерзляк ФГОС Задание 699

 

\[\boxed{\mathbf{699\ (699).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[4 = \sqrt{16}\]

\[3\sqrt{2} = \sqrt{18}\]

\[\sqrt{16};\sqrt{17};\sqrt{18} \Longrightarrow 4;\ \ \sqrt{17};\ \ \ \]

\[3\sqrt{2}.\]

\[\boxed{\mathbf{6}\mathbf{9}\mathbf{9}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ 4x + 4 = 3x² + 5x - 10\]

\[3x^{2} + x - 14 = 0\]

\[D = 1 + 4 \cdot 3 \cdot 14 = 169\]

\[x = \frac{- 1 \pm \sqrt{169}}{6} = \frac{- 1 \pm 13}{6}\]

\[x_{1} = - \frac{7}{3} = - 2\frac{1}{3}\]

\[x_{2} = 2\]

\[Ответ:\ x = - 2\frac{1}{3};x = 2.\ \]

\[2)\ 10p² + 10p + 8 =\]

\[= 3p² - 10p + 11\]

\[7p^{2} + 20p - 3 = 0\]

\[D = 400 + 4 \cdot 3 \cdot 7 = 484\]

\[p = \frac{- 20 \pm \sqrt{484}}{7 \cdot 2} = \frac{- 20 \pm 22}{14}\]

\[p_{1} = \frac{1}{7}\]

\[p_{2} = - 3\]

\[Ответ:x = \frac{1}{7};\ x = - 3.\]