Question

Решить:x^2+(x^2/(2x+1)^2)=6

Answer 1

Ответ:

$\frac{{ - 3 \pm \sqrt 3 }}{2};\ 1 \pm \sqrt 2$

Объяснение:

Сделаем замену

$t = x - \displaystyle\frac{x}{{2x + 1}} = \displaystyle\frac{{x(2x + 1) - x}}{{2x + 1}} = \displaystyle\frac{{2{x^2}}}{{2x + 1}}.$

Тогда

${t^2} = {\left( {x - \displaystyle\frac{x}{{2x + 1}}} \right)^2} = {x^2} - 2x \cdot \displaystyle\frac{x}{{2x + 1}} + \displaystyle\frac{{{x^2}}}{{{{(2x + 1)}^2}}} =\\\\= {x^2} + \displaystyle\frac{{{x^2}}}{{{{(2x + 1)}^2}}} - \displaystyle\frac{{2{x^2}}}{{2x + 1}} = {x^2} + \displaystyle\frac{{{x^2}}}{{{{(2x + 1)}^2}}} - t,$

откуда

${x^2} + \displaystyle\frac{{{x^2}}}{{{{(2x + 1)}^2}}} = {t^2} + t.$

Получаем уравнение

${t^2} + t = 6;\\\\{t^2} + t - 6 = 0;\\\\\left\{ \begin{array}{l}{t_1} + {t_2} = - 1,\\{t_1}{t_2} = - 6;\end{array} \right.\\\\{t_1} = - 3,\,\,{t_2} = 2.$

Делаем обратную замену:

1)

$\displaystyle\frac{{2{x^2}}}{{2x + 1}} = - 3;\\\\2{x^2} = - 6x - 3;\\\\2{x^2} + 6x + 3 = 0;\\\\D = {6^2} - 4 \cdot 2 \cdot 3 = 36 - 24 = 12;\\\\x = \displaystyle\frac{{ - 6 \pm 2\sqrt 3 }}{{2 \cdot 2}} = \displaystyle\frac{{ - 3 \pm \sqrt 3 }}{2}.$

2)

$\displaystyle\frac{{2{x^2}}}{{2x + 1}} = 2;\\\\2{x^2} = 4x + 2;\\\\2{x^2} - 4x - 2 = 0;\\\\{x^2} - 2x - 1 = 0;\\\\D = {( - 2)^2} - 4 \cdot ( - 1) = 4 + 4 = 8;\\\\x = \displaystyle\frac{{2 \pm 2\sqrt 2 }}{{2 \cdot 1}} = 1 \pm \sqrt 2 .$