Решебник по алгебре 8 класс Мерзляк Упражнения (страница 196)

\[\boxed{\mathbf{Упражнения}\mathbf{\ }\mathbf{стр}\mathbf{.\ 196}}\]

\[\boxed{\mathbf{1.}}\]

\[\frac{3x^{2} - 9x}{2} - \frac{12}{x^{2} - 3x} = 3\]

\[\frac{3 \cdot \left( x^{2} - 3x \right)}{2} - \frac{12}{x^{2} - 3x} = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \]

\[ОДЗ:\ \ x^{2} - 3x \neq 0\]

\[x(x - 3) \neq 0\]

\[x \neq 0;\ \ \ \ \ x \neq 3\]

\[Пусть\ \ y = x^{2} - 3x:\]

\[\frac{3y}{2} - \frac{12}{y} = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ | \cdot 2y\]

\[3y^{2} - 24 = 6y\]

\[3y^{2} - 6y - 24 = 0\]

\[y^{2} - 2y - 8 = 0\]

\[y_{1} + y_{2} = 2;\ \ \ \ \ \ y_{1} \cdot y_{2} = - 8\]

\[y_{1} = 4;\ \ \ \ \ \ \ \ y_{2} = - 2\]

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

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

\[x_{1} + x_{2} = 3;\ \ \ \ x_{1} \cdot x_{2} = - 4\]

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

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

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

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

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

\[Ответ:x = 1;\ x = - 1;x = 2;\]

\[x = 4.\]

\[\boxed{\mathbf{2.}}\]

\[ОДЗ:\ \ \ \ \ x \neq - 1;\ \ \ x \neq - 2;\ \ \ \ \]

\[x \neq 1;\ \ x \neq - 4.\]

\[(x + 1)(x + 2) =\]

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

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

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

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

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

\[\frac{6}{x^{2} + 3x + 2} + \frac{8}{x^{2} + 3x - 4} = 1\]

\[Пусть\ y = x^{2} + 3x:\]

\[\frac{6}{y + 2} + \frac{8}{y - 4} = 1\ \ \ \ \ \ \ \ \ \ \ y \neq - 2;\ \ \ \ \]

\[y \neq 4\]

\[6 \cdot (y - 4) + 8 \cdot (y + 2) =\]

\[= (y + 2)(y - 4)\]

\[6y - 24 + 8y + 16 =\]

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

\[14y - 8 - y^{2} + 2y + 8 = 0\]

\[- y^{2} + 16y = 0\]

\[y^{2} - 16y = 0\]

\[y(y - 16) = 0\]

\[y = 0;\ \ \ \ \ y = 16.\]

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

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

\[x = 0;\ \ \ \ x = - 3.\]

\[2)\ x^{2} + 3x = 16\]

\[{x^{2} + 3x - 16 = 0 }{D = 9 + 64 = 73}\]

\[x_{1,2} = \frac{- 3 \pm \sqrt{73}}{2}\]

\[Ответ:x = 0;\ x = - 3;\ \]

\[x = \ \frac{- 3 \pm \sqrt{73}}{2}.\]

\[\boxed{\mathbf{3.}}\]

\[x(x + 3)(x + 5)(x + 8) = 100\]

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

\[= 100\]

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

\[= 100\]

\[\left( x^{2} + 8x \right)\left( x^{2} + 8x + 15 \right) = 100\]

\[Пусть\ y = x^{2} + 8x:\]

\[y(y + 15) = 100\]

\[y^{2} + 15y - 100 = 0\]

\[y_{1} + y_{2} = - 15;\ \ \ \ \ y_{1} \cdot y_{2} = - 100\]

\[y_{1} = - 20;\ \ \ \ \ y_{2} = 5.\]

\[1)\ x^{2} + 8x + 20 = 0\]

\[D = 64 - 80 < 0\]

\[нет\ корней.\]

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

\[D = 64 + 20 = 84\]

\[x_{1,2} = \frac{- 8 \pm \sqrt{84}}{2} = - 4 \pm \sqrt{21}\]

\[Ответ:x = - 4 \pm \sqrt{21}.\]

\[\boxed{\mathbf{4.}}\]

\[(x + 2)(x + 3)(x + 8)(x + 12) =\]

\[= 4x^{2}\]

\[\frac{x^{2} + 14x + 24}{x} \cdot \frac{x^{2} + 11x + 24}{x} =\]

\[= 4\]

\[\left( x + 14 + \frac{24}{x} \right)\left( x + 11 + \frac{24}{x} \right) = 4\]

\[Пусть\ \ \ \ y = x + \frac{24}{x}:\]

\[(y + 14)(y + 11) = 4\]

\[y^{2} + 14y + 11y + 154 - 4 = 0\]

\[y^{2} + 25y + 150 = 0\]

\[y_{1} + y_{2} = - 25;\ \ \ \ \ \ y_{1} \cdot y_{2} = 150\]

\[y_{1} = - 15;\ \ \ \ \ \ y_{2} = - 10.\]

\[1)\ x + \frac{24}{x} = - 15\]

\[x^{2} + 24 = - 15x\]

\[x^{2} + 15x + 24 = 0\]

\[D = 225 - 96 = 129\]

\[x_{1,2} = \frac{- 15 \pm \sqrt{129}\ }{2}\]

\[2)\ x + \frac{24}{x} = - 10\]

\[x^{2} + 24 = - 10x\]

\[x^{2} + 10x + 24 = 0\]

\[x_{1} + x_{2} = - 10;\ \ \ \ \ x_{1} \cdot x_{2} = 24\]

\[x_{1} = - 6;\ \ \ \ \ \ x_{2} = - 4\]

\[Ответ:x = - 6;\ \ \ x = - 4;\ \ \ \ \]

\[x = \frac{- 15 \pm \sqrt{129}}{2}.\]

\[\boxed{\mathbf{5.}}\]

\[7 \cdot \left( x + \frac{1}{x} \right) - 2 \cdot \left( x^{2} + \frac{1}{x^{2}} \right) = 9;\ \ \ \ \ \ \ \ \ \]

\[x \neq 0\]

\[Пусть\ y = x + \frac{1}{x}:\]

\[y^{2} = \left( x + \frac{1}{x} \right)^{2} = x^{2} + 2 + \frac{1}{x^{2}}\]

\[x^{2} + \frac{1}{x^{2}} = y^{2} - 2\]

\[Подставим:\]

\[7y - 2 \cdot \left( y^{2} - 2 \right) = 9\]

\[7y - 2y^{2} + 4 - 9 = 0\]

\[- 2y^{2} + 7y - 5 = 0\]

\[2y^{2} - 7y + 5 = 0\]

\[D = 49 - 40 = 9\]

\[y_{1} = \frac{7 + 3}{4} = \frac{10}{4} = 2,5;\ \ \ \ \ \]

\[y_{2} = \frac{7 - 3}{4} = 1\]

\[1)\ x + \frac{1}{x} = 2,5\ \ \ \ \ \ \ \ | \cdot 2x\]

\[2x^{2} + 2 = 5x\]

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

\[D = 25 - 16 = 9\]

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

\[x_{2} = \frac{5 - 3}{4} = \frac{2}{4} = 0,5\]

\[2)\ x + \frac{1}{x} = 1\ \ \ \ \ \ \ \ | \cdot x\]

\[x^{2} + 1 = x\]

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

\[D = 1 - 4 = - 3 < 0\]

\[корней\ нет.\]

\[Ответ:x = 2;\ \ \ x = 0,5.\]

\[\boxed{\mathbf{6.}}\]

\[2 \cdot \left( x^{2} + x + 1 \right)^{2} - 7 \cdot (x - 1)^{2} =\]

\[= 13 \cdot \left( x^{3} - 1 \right)\]

\[2 \cdot \left( x^{2} + x + 1 \right)^{2} - 7 \cdot (x - 1)^{2} =\]

\[ОДЗ:\ \ \ \ x \neq 1.\]

\[x^{3} - 1 = (x - 1)\left( x^{2} + x + 1 \right)\]

\[\frac{2 \cdot \left( x^{2} + x + 1 \right)}{x - 1} - \frac{7 \cdot (x - 1)}{x^{2} + x + 1} =\]

\[= 13\]

\[Пусть\ \ y = \frac{x^{2} + x + 1}{x - 1}:\]

\[2y - \frac{7}{y} = 13\]

\[2y^{2} - 7 = 13y\]

\[2y^{2} - 13y - 7 = 0\]

\[D = 169 + 56 = 225\]

\[y_{1} = \frac{13 + 15}{4} = 7;\ \ \ \ \ \]

\[y_{2} = \frac{13 - 15}{4} = - \frac{2}{4} = - 0,5\]

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

\[x^{2} + x + 1 = 7 \cdot (x - 1)\]

\[x^{2} + x + 1 - 7x + 7 = 0\]

\[x^{2} - 6x + 8 = 0\]

\[x_{1} + x_{2} = 6;\ \ \ x_{1} \cdot x_{2} = 8\]

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

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

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

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

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

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

\[D = 9 - 8 = 1\]

\[x_{1} = \frac{- 3 + 1}{4} = - 0,5;\ \ \ \]

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

\[Ответ:x = 2;\ \ x = 4;\ \ x = - 1;\ \ \ \]

\[x = - 0,5.\]

\[\boxed{\mathbf{7.}}\]

\[(x - 6)^{4} + (x - 4)^{4} = 82\]

\[Пусть\ y = x - 5:\]

\[(y - 1)^{4} + (y + 1)^{4} = 82\]

\[2y^{4} + 12y^{2} + 2 = 82\]

\[2y^{4} + 12y^{2} - 80 = 0\ \ \ \ \ \ |\ :2\]

\[y^{4} + 6y^{2} - 40 = 0\]

\[Пусть\ y^{2} = z:\]

\[z^{2} + 6z - 40 = 0\]

\[z_{1} + z_{2} = - 6;\ \ \ \ z_{1} \cdot z_{2} = - 40\]

\[z_{1} = - 10;\ \ \ \ z_{2} = 4.\]

\[1)\ y^{2} = - 10\]

\[нет\ корней.\]

\[2)\ y^{2} = 4\]

\[y = \pm 2.\]

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

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

\[Ответ:x = 7;\ \ x = 3.\]