Question

9x(в 4)-13x(в квадрате)+4=0 (x(во второй)-2x)во второй + (х(во второй)-2х)=12

Answer 1

${9x}^{4} - 13 {x}^{2} + 4 = 0 \\$

Пусть

${x}^{2} = t$

тогда:

$9 {t}^{2} - 13t + 4 = 0 \\ d = 169 - 144 = 25 = {5}^{2} \\ t = \frac{13 - 5}{18} = \frac{8}{18} = \frac{4}{9} \\ t = \frac{13 + 5}{18} = \frac{18}{18} = 1$

Делаем обратную замену

${x }^{2} = \frac{4}{9} \\ x = \frac{2}{3} \\ x = - \frac{2}{3}$

${x}^{2} = 1 \\ x = 1 \\ x = - 1$

ответ: -1; -2/3; 2/3; 1

$( {x}^{2} - 2x) {}^{2} + ( {x}^{2} - 2x) = 12 \\$

Пусть

$( {x}^{2} - 2x) = t$

Тогда

$t {}^{2} + t - 12 = 0 \\ d = 1 + 48 = 49 = {7}^{2} \\ t = \frac{ - 1 - 7}{2} = \frac{ - 8}{2} = - 4 \\ t = \frac{ - 1 + 7}{2} = \frac{6}{2} = 3$

Делаем обратную замену

${ x}^{2} - 2x = - 4 \\ {x}^{2} - 2x + 4 = 0 \\ d = 4 - 16 = - 12 \\ d < 0 \\ net \: kornei$

${x}^{2} - 2x = 3 \\ {x}^{2} - 2x - 3 = 0 \\ d = 4 + 12 = 16 = {4}^{2} \\ x = \frac{2 - 4}{2} = \frac{ - 2}{2} = - 1 \\ x = \frac{2 + 4}{2} = \frac{6}{2} = 3$

ответ: -1;3