ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 106

\[\boxed{\mathbf{106}.}\]

\[1)\ 0,3x^{2} = 0\]

\[x^{2} = 0\]

\[x = 0.\]

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

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

\[x^{2} = - 0,02\]

\[нет\ корней.\]

\[3)\ x^{2} = 24\]

\[x = \pm \sqrt{24}\]

\[x = \pm 2\sqrt{6}.\]

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

\[x^{2} = 9\]

\[x = \pm 3.\]

\[5)\ \frac{1}{3}x^{2} + 6 = 0\]

\[\frac{1}{3}x^{2} = - 6\]

\[x^{2} = - 18\]

\[нет\ корней.\]

\[6) - x^{2} + \frac{1}{4} = 0\]

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

\[x = \pm \frac{1}{2} = \pm 0,5.\]

\[7)\ \frac{1}{5}x^{2} - 2x = 0\]

\[\frac{1}{5}x(x - 10) = 0\]

\[x = 0;\ \ x = 10.\]

\[8)\ 3x + 4x^{2} = 0\]

\[4x\left( x + \frac{3}{4} \right) = 0\]

\[x = 0;\ \ x = - 0,75.\]

\[9)\ x(x - 3) = 4 \cdot (x + 1) +\]

\[+ 3x^{2} - 7x\]

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

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

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

\[нет\ корней.\]

\[10)\ \frac{x^{2} - 2}{2} + \frac{2 + x^{2} - x}{3} =\]

\[= \frac{3x - 1}{3}\ \ \ \ | \cdot 6\]

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

\[= 6x - 2\]

\[5x^{2} - 8x = 0\]

\[5x\left( x - \frac{8}{5} \right) = 0\]

\[x = 0;\ \ x = 1,6.\]

\[11)\ x^{2} - 7x + 12 = 0\]

\[x_{1} + x_{2} = 7;\ \ \ x_{1} \cdot x_{2} = 12\]

\[x_{1} = 3;\ \ x_{2} = 4.\]

\[12)\ x^{2} + x - 30 = 0\]

\[x_{1} + x_{2} = - 1;\ \ \ x_{1} \cdot x_{2} = - 30\]

\[x_{1} = - 6;\ \ x_{2} = 5.\]

\[13)\ x^{2} + 4x + 9 = 0\]

\[D_{1} = 4 - 9 = - 5 < 0\]

\[нет\ корней.\]

\[14)\ x^{2} + 3x - 108 = 0\]

\[x_{1} \cdot x_{2} = - 3;\ \ x_{1} \cdot x_{2} = - 108\]

\[x_{1} = - 12;\ \ x_{2} = 9.\]

\[15)\ x^{2} + 2\sqrt{3}x + 3 = 0\]

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

\[x + \sqrt{3} = 0\]

\[x = - \sqrt{3}.\]

\[16)\ \frac{1}{4}x^{2} - 2x + 4 = 0\]

\[\left( \frac{1}{2}x - 2 \right)^{2} = 0\]

\[\frac{1}{2}x - 2 = 0\]

\[\frac{1}{2}x = 2\]

\[x = 4.\]

\[17)\ 2x^{2} + x - 15 = 0\]

\[D = 1 + 120 = 121\]

\[x_{1} = \frac{- 1 + 11}{4} = \frac{10}{4} = 2,5;\ \ \ \]

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

\[18)\ 3x^{2} - 14x + 8 = 0\]

\[D_{1} = 49 - 24 = 25\]

\[x_{1} = \frac{7 + 5}{3} = 4;\ \ \ x_{2} = \frac{7 - 5}{3} = \frac{2}{3}.\]

\[19) - 4x^{2} + 11x + 3 = 0\]

\[4x^{2} - 11x - 3 = 0\]

\[D = 121 + 48 = 169\]

\[x_{1} = \frac{11 + 13}{8} = 3;\ \ \ \]

\[x_{2} = \frac{11 - 13}{8} = - \frac{2}{8} =\]

\[= - \frac{1}{4} = - 0,25.\]

\[20) - 2x^{2} + 3x - 3 = 0\]

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

\[D = 9 - 24 = - 15 < 0\]

\[нет\ корней.\]