[tex]\displaystyle \sf 2022^x+\sqrt{2022^x} = 2023 \\\\\ \text{Fa\c camos }\\\\ 2022^x= a \ \ ;\ \ a > 0 \\\\ Da{\'i}}: \\\\ a+\sqrt{a}=2023 \\\\ \underbrace{\sf \sqrt{a}}_{ > 0}=\underbrace{\sf 2023-a}_{ > 0} \\\\ 2023-a > 0 \\\\ a < 2023 \to 2022^x < 2023 \\\\ \boxed{\sf 1 < x < 2 } \to \text{Com essa informa\c c\~ao vamos nos atentar}\\\\ \text{ temos }:[/tex]

[tex]\displaystyle \sf (\sqrt{a})^2=(2023-a)^2 \\\\ a=2023^2-2a\cdot 2023+a^2 \\\\ a^2-2a\cdot 2023-a+2023^2 = 0 \\\\ a^2-(2\cdot 2023+1)\cdot a +2023^2 = 0 \\\\ a = \frac{-[-(2\cdot 2023+1)]\pm\sqrt{(2\cdot 2023+1)^2-4\cdot 2023^2}}{2}[/tex]

[tex]\displaystyle \sf a= \frac{2\cdot 2023+1\pm\sqrt{4\cdot 2023^2+2\cdot 2\cdot 2023\cdot 1+1-4\cdot 2023^2}}{2} \\\\\\ a = \frac{4047\pm\sqrt{4\cdot 2023+1}}{2} \to a= \frac{4047\pm\sqrt{8093}}{2}[/tex]

Daí :

[tex]\displaystyle \sf 2022^x=\frac{4047\pm\sqrt{8093}}{2} \\\\\ \text{J\'a que }1 < x < 2 \text{ \ e } \sqrt{8093} < 4047 \text{\ (Ent\~ao vamos usar a parte negativa)} \\\\\ temos : \\\\ 2022^x=\frac{4047-\sqrt{8093}}{2} \\\\\ ln(2022)^x =\ln\left(\frac{4047-\sqrt{8093}}{2}\right) \\\\\ x \cdot \ln 2022 = \ln\left(\frac{4047-\sqrt{8093}}{2}\right) \\\\\\ \Large\boxed{\sf x=\ln\left(\frac{4047-\sqrt{8093}}{2}\right) \cdot \left[ \ \ln(2022) \ \right]^{-1}\ }\checkmark[/tex]

OU

[tex]\displaystyle \sf \Large\boxed{\sf \ x =\frac{\left[\ \ln (4047-\sqrt{8093})-\ln(2)\ \right] }{\ln(2022)}\ }\checkmark[/tex]

Verified answer

Resposta:

[tex]\textsf{Leia abaixo}[/tex]

Explicação passo a passo:

[tex]\sf (2022)^x + \sqrt{(2022)^x} = 2023[/tex]

[tex]\sf y = \sqrt{(2022)^x}[/tex]

[tex]\sf y^2 + y - 2023 = 0[/tex]

[tex]\sf a = 1 \Leftrightarrow b = 1 \Leftrightarrow c = -2023[/tex]

[tex]\sf \Delta = b^2 - 4.a.c[/tex]

[tex]\sf \Delta = (1)^2 - 4.1.(-2023)[/tex]

[tex]\sf \Delta = 1 + 8092[/tex]

[tex]\sf \Delta = 8093[/tex]

[tex]\sf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{-1 \pm \sqrt{8093}}{2} \rightarrow \begin{cases}\sf{x' = \dfrac{-1 + \sqrt{8093}}{2}}\\\\\sf{x'' = \dfrac{-1 - \sqrt{8093}}{2}}\end{cases}}[/tex]

[tex]\sf (\sqrt{2022})^x = \dfrac{-1 + \sqrt{8093}}{2}[/tex]

[tex]\boxed{\sf x = log_{\:2022}\left(\dfrac{-1 + \sqrt{8093}}{2}\right)^2}[/tex]