Question

РЕШИТЬ ПОКАЗАТЕЛЬНОЕ УРАВНЕНИЕС РЕШЕНИЕМ">

Answer 1

$5^{3x^2 + 3x - 4} - 6\cdot 25^{x^2 + x - 1} + 5^{x^2 + x + 1} = 0\\\\5^{3x^2 + 3x - 4} - 6\cdot 5^{2x^2 + 2x - 2} + 5^{x^2 + x + 1} = 0\\\\5^{x^2 + x + 1}\left(\dfrac{5^{3x^2 + 3x - 4}}{5^{x^2 + x + 1}} - 6\cdot\dfrac{5^{2x^2 + 2x - 2}}{5^{x^2+x + 1}} + 1\right) = 0\\\\\\5^{x^2 + x + 1}\left(5^{3x^2 + 3x - 4 - \left(x^2 + x + 1\right)} - 6\cdot 5^{2x^2 + 2x - 2 - \left(x^2 + x +1\right)} + 1\right) = 0\\\\5^{x^2 + x +1}\left(5^{2x^2 + 2x-5} - 6\cdot 5^{x^2 + x - 3} + 1\right) = 0$

$5^{x^2 + x + 1}\neq 0$, поэтому делим обе части уравнения.

$5^{2x^2 + 2x - 5} - 6\cdot 5^{x^2 + x - 3} + 1 = 0$

Берём замену $t = 5^{x^2 + x - 3}\ ,t0$. Тогда: $5^{2x^2 + 2x - 5} = \left(5^{x^2 + x - 3}\right)^2\cdot 5 = 5t^2$ .

$5t^2 - 6t + 1 = 0\\\\D = b^2 - 4ac = 36 - 20 = 16\\\\t_1 = \dfrac{-b+\sqrt{D}}{2a} = \dfrac{6 + 4}{10} = \boldsymbol{1}\\\\\\t_2 = \dfrac{-b-\sqrt{D}}{2a} = \dfrac{6-4}{10} = \boldsymbol{\dfrac{1}{5}}$

Обратная замена.

$\left[\begin{gathered}5^{x^2 + x - 3} = 1\\\\5^{x^2+x-3} = \dfrac{1}{5}\end{gathered}\ \ \ \ \Leftrightarrow\ \left[\begin{gathered}5^{x^2 + x - 3} = 5^0\\\\5^{x^2+x-3} = 5^{-1}\end{gathered}\ \ \ \ \Leftrightarrow\ \left[\begin{gathered}x^2 + x - 3 = 0\\x^2 + x - 3 = -1\end{gathered}$

Решим уравнения отдельно.

1)

$x^2 + x - 3 = 0\\\\D = b^2 - 4ac = 1 + 12 = 13\\\\x_1 = \dfrac{-b+\sqrt{D}}{2a}= \dfrac{-1 + \sqrt{13}}{2} = \boldsymbol{\dfrac{\sqrt{13} - 1}{2}}\\\\\\x_2 = \dfrac{-b-\sqrt{D}}{2a} = \boldsymbol{\dfrac{-1-\sqrt{13}}{2}}$

2)

$x^2 + x - 3 = -1\\\\x^2 + x - 2 = 0\\\\\begin{equation*}\begin{cases}x_1x_2 = -2\\x_1 + x_2 = -1\end{cases}\end{equation*}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big|\boldsymbol{x = -2;\ x = 1}$

Итого имеем 4 корня:

$$\left[\begin{gathered}x = \dfrac{\sqrt{13} - 1}{2}\\\\x = \dfrac{-1-\sqrt{13}}{2}\\\\x = -2\\\\x = 1\end{gathered}$

ответ:   $\dfrac{-1 - \sqrt{13}}{2};\ -2\,;\ 1\,;\ \dfrac{\sqrt{13} - 1}{2}\ .$