Giải PT: 2022$\sqrt[]{2022x-2021}$`+`$\sqrt[]{2023x-2022}$ =2023 11/07/2021 Bởi Rose Giải PT: 2022$\sqrt[]{2022x-2021}$`+`$\sqrt[]{2023x-2022}$ =2023
Điều kiện xác định $x\ge \dfrac{2022}{2023}$ $\begin{array}{l} 2022\sqrt {2022x – 2021} + \sqrt {2023x – 2022} = 2023\\ \Leftrightarrow 2022\sqrt {2022x – 2021} – 2022 + \sqrt {2023x – 2022} – 1 = 0\\ \Leftrightarrow 2022\left( {\sqrt {2022x – 2021} – 1} \right) + \left( {\sqrt {2023x – 2022} – 1} \right) = 0\\ \Leftrightarrow 2022.\dfrac{{2022x – 2022}}{{\sqrt {2022x – 2021} + 1}} + \dfrac{{2023x – 2023}}{{\sqrt {2023x – 2022} + 1}} = 0\\ \Leftrightarrow \left( {x – 1} \right)\left( {\underbrace {\dfrac{{{{2022}^2}}}{{\sqrt {2022x – 2021} + 1}} + \dfrac{{2023}}{{\sqrt {2023x – 2022} + 1}}}_{ > 0}} \right) = 0\\ \Leftrightarrow x = 1\\ \Rightarrow S = \left\{ 1 \right\} \end{array}$ Bình luận
Điều kiện xác định $x\ge \dfrac{2022}{2023}$
$\begin{array}{l} 2022\sqrt {2022x – 2021} + \sqrt {2023x – 2022} = 2023\\ \Leftrightarrow 2022\sqrt {2022x – 2021} – 2022 + \sqrt {2023x – 2022} – 1 = 0\\ \Leftrightarrow 2022\left( {\sqrt {2022x – 2021} – 1} \right) + \left( {\sqrt {2023x – 2022} – 1} \right) = 0\\ \Leftrightarrow 2022.\dfrac{{2022x – 2022}}{{\sqrt {2022x – 2021} + 1}} + \dfrac{{2023x – 2023}}{{\sqrt {2023x – 2022} + 1}} = 0\\ \Leftrightarrow \left( {x – 1} \right)\left( {\underbrace {\dfrac{{{{2022}^2}}}{{\sqrt {2022x – 2021} + 1}} + \dfrac{{2023}}{{\sqrt {2023x – 2022} + 1}}}_{ > 0}} \right) = 0\\ \Leftrightarrow x = 1\\ \Rightarrow S = \left\{ 1 \right\} \end{array}$
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