Решебник по алгебре 8 класс Мерзляк ФГОС Задание 706

 

\[\boxed{\mathbf{706\ (706).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[x^{2} - 14x + 12 = 0\]

\[x_{1}x_{2} = \frac{c}{a} = 12\]

\[Ответ:3)\ 12.\]

\[\boxed{\mathbf{706.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ (3x - 1)(x + 4) = (2x + 3)(x + 3) - 17\]

\[3x^{2} - x + 12x - 4 = 2x^{2} + 3x + 6x + 9 - 17\]

\[x^{2} + 11x - 9x - 4 + 8 = 0\]

\[x^{2} + 2x + 4 = 0\]

\[(x + 2)^{2} = 0\]

\[x = - 2.\]

\[Ответ:x = - 2.\]

\[2)\ x^{2} + 3x\sqrt{2} + 4 = 0\]

\[D = 18 - 16 = 2\]

\[x_{1} = \frac{\left( 3\sqrt{2} + \sqrt{2} \right)}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2};\]

\[x_{2} = \frac{3\sqrt{2} - \sqrt{2}}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2}.\]

\[Ответ:2\sqrt{2};\ \ \sqrt{2}.\]

\[3)\ x^{2} - x\left( \sqrt{3} + 2 \right) + 2\sqrt{3} = 0\]

\[D = \left( \sqrt{3} + 2 \right)^{2} - 4 \cdot 2\sqrt{3} =\]

\[= 3 + 4\sqrt{3} + 4 - 8\sqrt{3} =\]

\[= 3 - 4\sqrt{3} + 4 = \left( \sqrt{3} - 2 \right)^{2}\]

\[x_{1} = \frac{\sqrt{3} + 2 + \sqrt{3} - 2}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3};\]

\[x_{2} = \frac{\sqrt{3} + 2 - \sqrt{3} + 2}{2} = \frac{4}{2} = 2.\]

\[Ответ:\sqrt{3};2.\]

\[4)\ \frac{2x^{2} + x}{3} - \frac{x + 3}{4} = x - 1\ \ \ \ | \cdot 12\]

\[4\left( 2x^{2} + x \right) - 3(x + 3) = 12(x - 1)\]

\[8x^{2} + 4x - 3x - 9 = 12x - 12\]

\[8x^{2} - 11x + 3 = 0\]

\[D = 121 - 96 = 25\]

\[x_{1} = \frac{11 + 5}{16} = 1;\]

\[x_{2} = \frac{11 - 5}{16} = \frac{6}{16} = \frac{3}{8}.\]

\[Ответ:1;\ \ \ \frac{3}{8}.\]