ГДЗ по алгебре 8 класс Мерзляк Задание 628

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

\[3x^{2} + 12x - x - 4 + 4 = 0\]

\[3x² + 11x = 0\]

\[x(3x + 11) = 0\]

\[x = 0,\ \ \ \ 3x + 11 = 0\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = - \frac{11}{3} = - 3\frac{2}{3}\]

\[Ответ:\ x = - 3\frac{2}{3};0.\]

\[2)\ (2x - 1)^{2} - 6 \cdot (6 - x) = 2x\]

\[4x^{2} - 4x + 1 - 36 + 6x - 2x =\]

\[= 0\]

\[4x^{2} - 6x + 6x - 35 = 0\]

\[4x^{2} = 35\]

\[x^{2} = \frac{35}{4}\]

\[x = \sqrt{\frac{35}{4}}\]

\[x = - \sqrt{\frac{35}{4}}\]

\[x = \frac{\sqrt{35}}{2}\]

\[x = - \frac{\sqrt{35}}{2}\]

\[Ответ:\ x = - \frac{\sqrt{35}}{2};\frac{\sqrt{35}}{2}.\]

\[x^{2} - x - 6 - x^{2} + 25 - x^{2} + x =\]

\[= 0\]

\[- x^{2} + 19 = 0\]

\[x^{2} = 19\]

\[x = \sqrt{19}\]

\[x = - \sqrt{19}\]

\[Ответ:\ x = - \sqrt{19};\sqrt{19}.\]