Giải PT: 2022$\sqrt[]{2022x-2021}$`+`$\sqrt[]{2023x-2022}$ =2023

Điều kiện xác định $x\ge \dfrac{2022}{2023}$

$\begin{array}{l} 2022\sqrt {2022x – 2021}  + \sqrt {2023x – 2022}  = 2023\\  \Leftrightarrow 2022\sqrt {2022x – 2021}  – 2022 + \sqrt {2023x – 2022}  – 1 = 0\\  \Leftrightarrow 2022\left( {\sqrt {2022x – 2021}  – 1} \right) + \left( {\sqrt {2023x – 2022}  – 1} \right) = 0\\  \Leftrightarrow 2022.\dfrac{{2022x – 2022}}{{\sqrt {2022x – 2021}  + 1}} + \dfrac{{2023x – 2023}}{{\sqrt {2023x – 2022}  + 1}} = 0\\  \Leftrightarrow \left( {x – 1} \right)\left( {\underbrace {\dfrac{{{{2022}^2}}}{{\sqrt {2022x – 2021}  + 1}} + \dfrac{{2023}}{{\sqrt {2023x – 2022}  + 1}}}_{ > 0}} \right) = 0\\  \Leftrightarrow x = 1\\  \Rightarrow S = \left\{ 1 \right\} \end{array}$