Решебник по алгебре 8 класс Мерзляк Задание 754

\[\boxed{\mathbf{754\ (754).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ x² - 3x - 18 = 0\]

\[x_{1} + x_{2} = 3,\ \ x_{1} = - 3\]

\[x_{1} \cdot x_{2} = - 18,\ \ x_{2} = 6\]

\[x² - 3x - 18 = (x + 3)(x - 6)\]

\[2)\ x² + 5x - 14 = 0\]

\[x_{1} + x_{2} = - 5,\ \ x_{1} = - 7\]

\[x_{1} \cdot x_{2} = 14,\ \ x_{2} = 2\]

\[x^{2} + 5x - 14 = (x + 7)(x - 2)\]

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

\[x_{1} + x_{2} = 3,\ \ x_{1} = 4\]

\[x_{1} \cdot x_{2} = - 4,\ \ x_{2} = - 1\]

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

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

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

\[4)\ 5x² + 8x - 4 = 0\]

\[x_{1} + x_{2} = - \frac{8}{5},\ \ \]

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

\[x_{1} \cdot x_{2} = - \frac{4}{5},\ \ x_{2} = \frac{2}{5}\ \]

\[5x^{2} + 8x - 4 =\]

\[= 5 \cdot (x + 2)\left( x - \frac{2}{5} \right) =\]

\[= (x + 2)(5x - 2)\]

\[5)\ 2a² - 3a + 1 = 0\]

\[a_{1} + a_{2} = \frac{3}{2},\ \ a_{1} = \frac{2}{2} = 1\]

\[a_{1} \cdot a_{2} = \frac{1}{2},\ \ a_{2} = \frac{1}{2}\]

\[2a^{2} - 3a + 1 =\]

\[= 2 \cdot (a - 1)\left( a - \frac{1}{2} \right) =\]

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

\[6)\ 4b² - 11b - 3 = 0\]

\[b_{1} + b_{2} = \frac{11}{4},\ \ b_{1} = - \frac{1}{4}\]

\[b_{1} \cdot b_{2} = - \frac{3}{4},\ \ b_{2} = \frac{12}{4} = 3\]

\[4b^{2} - 11b - 3 =\]

\[= 4 \cdot \left( b + \frac{1}{4} \right)(b - 3) =\]

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

\[7) - \frac{1}{4}x^{2} - 2x - 3 = 0\]

\[x_{1} + x_{2} = - 8,\ \ x_{1} = - 2\]

\[x_{1} \cdot x_{2} = 12,\ \ x_{2} = - 6\]

\[- \frac{1}{4}x^{2} - 2x - 3 =\]

\[= - \frac{1}{4} \cdot (x + 2)(x + 6)\]

\[8)\ 0,3m² - 3m + 7,5 = 0\]

\[m_{1} + m_{2} = \frac{3}{0,3} = 10,\ \ m_{1} = 5\]

\[m_{1} \cdot m_{2} = \frac{7,5}{0,3} = 25,\ \ m_{2} = 5\]

\[0,3m^{2} - 3m + 7,5 =\]

\[= 0,3 \cdot (m - 5)²\]

\[9)\ x² - 2x - 2 = 0\]

\[D = 4 + 8 = 12\]

\[x = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3}\]

\[x^{2} - 2x - 2 =\]

\[= (x - 1 - \sqrt{3})(x + 1 - \sqrt{3})\]