ГДЗ по алгебре 8 класс Макарычев Задание 540

\[\boxed{\text{540\ (540).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 8x^{2} - 14x + 5 = 0\]

\[D_{1} = 7^{2} - 8 \cdot 5 = 49 - 40 = 9\]

\[x_{1,2} = \frac{7 \pm \sqrt{9}}{8} = \frac{7 \pm 3}{8}\]

\[x_{1} = \frac{10}{8} = \frac{5}{4} = 1,25;\ \ \]

\[x_{2} = \frac{4}{8} = \frac{1}{2} = 0,5\]

\[Ответ:x = 1,25;\ \ x = 0,5.\]

\[\textbf{б)}\ 12x^{2} + 16x - 3 = 0\]

\[D_{1} = 8^{2} + 3 \cdot 12 = 64 + 36 =\]

\[= 100\]

\[x_{1,2} = \frac{- 8 \pm \sqrt{100}}{12} = \frac{- 8 \pm 10}{12}\]

\[x_{1} = - \frac{18}{12} = - \frac{3}{2} = - 1,5;\ \ \]

\[x_{2} = \frac{2}{12} = \frac{1}{6}\]

\[Ответ:x = - 1,5;\ \ x = \frac{1}{6}.\]

\[\textbf{в)}\ 4x^{2} + 4x + 1 = 0\]

\[D_{1} = 2^{2} - 1 \cdot 4 = 4 - 4 = 0\]

\[x = - \frac{2}{4} = - 0,5\]

\[Ответ:x = - 0,5.\]

\[\textbf{г)}\ x^{2} - 8x - 84 = 0\]

\[D_{1} = 4^{2} + 1 \cdot 84 = 16 + 84 =\]

\[= 100\]

\[x_{1,2} = \frac{4 \pm \sqrt{100}}{1} = 4 \pm 10\]

\[x_{1} = 14;\ \ x_{2} = - 6\]

\[Ответ:x = - 6;\ \ x = 14.\]

\[\textbf{д)}\ x^{2} + 6x - 19 = 0\]

\[D_{1} = 3^{2} + 1 \cdot 19 = 9 + 19 =\]

\[= 28 = 2\sqrt{7}\]

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

\[Ответ:x = - 3 \pm 2\sqrt{7}.\]

\[\textbf{е)}\ 5x^{2} + 26x - 24 = 0\]

\[D_{1} = 13^{2} + 5 \cdot 24 =\]

\[= 169 + 120 = 289\]

\[x_{1,2} = \frac{- 13 \pm \sqrt{289}}{5} = \frac{- 13 \pm 17\ }{5}\]

\[x_{1} = - 6;\ \ x_{2} = \frac{4}{5} = 0,8\]

\[Ответ:x = - 6;\ \ x = 0,8.\]

\[\textbf{ж)}\ x^{2} - 34x + 289 = 0\]

\[D_{1} = 17^{2} - 1 \cdot 289 =\]

\[= 289 - 289 = 0\]

\[x = \frac{17}{1} = 17\]

\[Ответ:x = 17.\]

\[\textbf{з)}\ 3x^{2} + 32x + 80 = 0\]

\[D_{1} = 16^{2} - 3 \cdot 240 =\]

\[= 256 - 240 = 16\]

\[x_{1,2} = \frac{- 16 \pm \sqrt{16}}{3} = \frac{- 16 \pm 4}{3}\]

\[x_{1} = - \frac{20}{3} = - 6\frac{2}{3};\ \ \]

\[x_{2} = - \frac{12}{3} = - 4\ \ \]

\[Ответ:x = - 6\frac{2}{3};\ \ x = - 4.\]