## Sum the solutions of the equation ({2021^{2021x}} + {2021^{x + 1}}.x = {2021^{{x^2}}} + {2021^x}. {x^2 }). – Math book

Sum the solutions of the equation $${2021^{2021x}} + {2021^{x + 1}}.x = {2021^{{x^2}}} + {2021^x}. {x^ 2}$$.

A. $$2.$$ B. $$2021.$$ C. $$2022.$$ D. $$2023.$$

Condition: $$x \in \mathbb{R}$$
Divide both sides of the equation by $${2021^x} > 0$$, we get: $${2021^{2020x}} + 2021x = {2021^{{x^2} – x}} + {x ^2}$$
$$\Leftrightarrow {2021^{2020x}} + 2020x = {2021^{{x^2} – x}} + {x^2} – x$$
Consider the function $$f\left( t \right) = {2021^t} + t$$ on $$\mathbb{R}$$
$$f’\left( t \right) = {2021^t}.\ln \left( {2021} \right) + 1 > 0,\,\forall t$$
$$\Rightarrow$$ function $$f\left( t \right)$$ covariates on $$\mathbb{R}$$
where $$f\left( {2021x} \right) = f\left( {{x^2} – x} \right)$$
From this it follows: $$2021x = {x^2} – x \Leftrightarrow \left[\begin{array}{l}x=0\\x=2022\end{array}\right$$[\begin{array}{l}x=0\x=2022\end{array}\right\)
So we have: $${x_1} + {x_2} = 2022$$.