ГДЗ по алгебре 8 класс Мерзляк Задание 753

\[\boxed{\mathbf{753\ (753).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ x² - 7x + 12 = 0\]

\[x_{1} + x_{2} = 7,\ \ x_{1} = 4\]

\[x_{1} \cdot x_{2} = 12,\ \ x_{2} = 3\]

\[x^{2} - 7x + 12 = (x - 4)(x - 3)\]

\[2)\ x² + 8x + 15 = 0\]

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

\[x_{1} \cdot x_{2} = 15,\ \ x_{2} = - 3\]

\[x^{2} + 8x + 15 = (x + 5)(x + 3)\]

\[3)\ x² - 3x - 10 = 0\]

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

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

\[x² - 3x - 10 = (x + 2)(x - 5)\]

\[4) - x^{2} - 5x - 6 = 0\]

\[x_{1} + x_{2} = - 5,\ \ x_{1} = - 3\]

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

\[- x^{2} - 5x - 6 = - (x + 3)(x + 2)\]

\[5) - x^{2} + x + 2 = 0\]

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

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

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

\[6)\ 6x² - 5x - 1 = 0\]

\[x_{1} + x_{2} = \frac{5}{6},\ \ x_{1} = - \frac{1}{6}\]

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

\[6x^{2} - 5x - 1 =\]

\[= 6 \cdot \left( x + \frac{1}{6} \right)(x - 1) =\]

\[= (6x + 1)(x - 1)\]

\[7)\ 4x² + 3x - 22 = 0\]

\[x_{1} + x_{2} = - \frac{3}{4},\ \ x_{1} = - \frac{11}{4}\]

\[x_{1} \cdot x_{2} = - \frac{22}{4},\ \ x_{2} = \frac{8}{4} = 2\]

\[4x^{2} + 3x - 22 =\]

\[= 4 \cdot \left( x + \frac{11}{4} \right)(x - 2) =\]

\[= (4x + 11)(x - 2)\]

\[8) - 3a^{2} + 8a + 3 = 0\]

\[a_{1} + a_{2} = \frac{8}{3},\ \ a_{1} = \frac{9}{3} = 3\]

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

\[- 3a^{2} + 8a + 3 =\]

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

\[= (3 - a)(3a + 3)\]

\[9)\ \frac{1}{6}b² - \frac{5}{6}b + 1 = 0\]

\[b_{1} + b_{2} = \frac{5 \cdot 6}{6 \cdot 1} = 5,\ \ b_{1} = 2\]

\[b_{1} \cdot b_{2} = 6,\ \ b_{2} = 3\]

\[\frac{1}{6}b^{2} - \frac{5}{6}b + 1 =\]

\[= \frac{1}{6} \cdot (b - 2)(b - 3)\]

\[10) - 2x^{2} - 0,5x + 1,5 = 0\]

\[x_{1} + x_{2} = - 0,25,\ \ x_{1} = 0,75\]

\[x_{1} \cdot x_{2} = - 0,75,\ \ x_{2} = - 1\]

\[- 2x^{2} - 0,5x + 1,5 =\]

\[= - 2 \cdot (x - 0,75)(x + 1) =\]

\[= (x + 1)(1,5 - 2x)\]

\[11)\ 0,4x² - 2x + 2,5 = 0\]

\[D = 4 - 4 = 0\]

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

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

\[= 0,4 \cdot (x - 2,5)^{2}\]

\[12) - 1,2m^{2} + 2,6m - 1 = 0\]

\[D = 6,76 - 4,8 = 1,96\]

\[x_{1} = \frac{- 2,6 - 1,4}{- 2,4} = \frac{5}{3}\]

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

\[- 1,2m^{2} + 2,6m - 1 =\]

\[= - 1,2 \cdot \left( m - \frac{5}{3} \right)(m - 0,5) =\]

\[= (2 - 1,2m)(m - 0,5)\]