Question

решите 2 уравнения, срочно!!! ​

Answer 1

Ответ:

$\dfrac{3}{1+x+x^{2} } =3-x-x^{2}$

$\dfrac{3}{1+x+x^{2} } (1+x+x^{2} )=3(1+x+x^{2} )-x(1+x+x^{2} )-x^{2} (1+x+x^{2} )$

$3=3(1+x+x^{2} )-x(1+x+x^{2} )-x^{2} (1+x+x^{2} )$

$x_{1} =0$     $x_{2} =2$     $x_{3} =-1$    $x_{4} =1$

$\dfrac{3x^{2} -9x}{2} -\dfrac{12}{x^{2} -3x} =3$

$x(3x^{2} -9x)(x-3)-24=6x(3-x)$

$x_{1} =-1$     $x_{2} =1$     $x_{3} =2$    $x_{4} =4$

Answer 2

Ответ:

13. –2, –1, 0, 1; 14. –2, 1, 2, 4.

Пошаговое объяснение:

13. Сделаем замену ${x^2} + x = t.$ Тогда

$\frac{3}{{1 + t}} = 3 - t;\\\\(3 - t)(1 + t) = 3;\\\\3 + 2t - {t^2} = 3;\\\\{t^2} - 2t = 0;\\\\t(t - 2) = 0;\\\\{t_1} = 0;\,\,{t_2} = 2.$

Делая обратную замену, получаем

${x^2} + x = 0;\\\\x(x + 1) = 0;\\\\{x_1} = 0;\,\,{x_2} = - 1;$

${x^2} + x = 2;\\\\{x^2} + x - 2 = 0;\\\\\left\{ \begin{array}{l}{x_3} + {x_4} = - 1,\\{x_3}{x_4} = - 2;\end{array} \right.\\\\{x_3} = - 2;\,\,{x_4} = 1.$

14. Сделаем замену ${x^2} - 3x = t.$ Тогда

$\frac{{3t}}{2} - \frac{{12}}{t} = 3;\\\\3{t^2} - 6t - 24 = 0;\\\\{t^2} - 2t - 8 = 0;\\\\\left\{ \begin{array}{l}{t_1} + {t_2} = 2,\\{t_1}{t_2} = - 8;\end{array} \right.\\\\{t_1} = - 2;\,\,{t_2} = 4.$

Делая обратную замену, получаем

${x^2} - 3x = - 2;\\\\{x^2} - 3x + 2 = 0;\\\\\left\{ \begin{array}{l}{x_1} + {x_2} = 3,\\{x_1}{x_2} = 2;\end{array} \right.\\\\{x_1} = 1;\,\,{x_2} = 2;$

${x^2} - 3x = 4;\\\\{x^2} - 3x - 4 = 0;\\\\\left\{ \begin{array}{l}{x_3} + {x_4} = 3,\\{x_3}{x_4} = - 4\end{array} \right.\\\\{x_3} = - 1;\,\,{x_4} = 4.$