Đáp án đúng: D
Giải chi tiết:Điều kiện: \(\left\{ \begin{array}{l}x > 2019\\y > 2020\\z > 2021\end{array} \right..\)
\(\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\dfrac{{1 - \sqrt {x - 2019} }}{{x - 2019}} + \dfrac{{1 - \sqrt {y - 2020} }}{{y - 2020}} + \dfrac{{1 - \sqrt {z - 2021} }}{{z - 2021}} + \dfrac{3}{4} = 0\\ \Leftrightarrow \dfrac{{1 - \sqrt {x - 2019} }}{{x - 2019}} + \dfrac{1}{4} + \dfrac{{1 - \sqrt {y - 2020} }}{{y - 2020}} + \dfrac{1}{4} + \dfrac{{1 - \sqrt {z - 2021} }}{{z - 2021}} + \dfrac{1}{4} = 0\\ \Leftrightarrow \dfrac{{4 - 4\sqrt {x - 2019} + \left( {x - 2019} \right)}}{{4\left( {x - 2019} \right)}} + \dfrac{{4 - 4\sqrt {y - 2020} + \left( {y - 2020} \right)}}{{4\left( {y - 2020} \right)}} + \dfrac{{4 - 4\sqrt {z - 2021} + \left( {z - 2021} \right)}}{{4\left( {z - 2021} \right)}} = 0\\ \Leftrightarrow \dfrac{{{{\left( {\sqrt {x - 2019} - 2} \right)}^2}}}{{4\left( {x - 2019} \right)}} + \dfrac{{{{\left( {\sqrt {y - 2020} - 2} \right)}^2}}}{{4\left( {y - 2020} \right)}} + \dfrac{{{{\left( {\sqrt {z - 2021} - 2} \right)}^2}}}{{4\left( {z - 2021} \right)}} = 0\,\,\,\left( * \right)\end{array}\)
Ta có: \(\left\{ \begin{array}{l}\dfrac{{{{\left( {\sqrt {x - 2019} - 2} \right)}^2}}}{{4\left( {x - 2019} \right)}} \ge 0\,\,\,\forall x > 2019\\\dfrac{{{{\left( {\sqrt {y - 2020} - 2} \right)}^2}}}{{4\left( {y - 2020} \right)}} \ge 0\,\,\forall y > 2020\\\dfrac{{{{\left( {\sqrt {z - 2021} - 2} \right)}^2}}}{{4\left( {z - 2021} \right)}} \ge 0\,\,\forall z > 2021\end{array} \right.\)
\( \Rightarrow \left( * \right) \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{{{\left( {\sqrt {x - 2019} - 2} \right)}^2}}}{{4\left( {x - 2019} \right)}} = 0\,\,\\\dfrac{{{{\left( {\sqrt {y - 2020} - 2} \right)}^2}}}{{4\left( {y - 2020} \right)}} = 0\,\\\dfrac{{{{\left( {\sqrt {z - 2021} - 2} \right)}^2}}}{{4\left( {z - 2021} \right)}} = 0\,\,\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\sqrt {x - 2019} - 2 = 0\\\sqrt {y - 2020} - 2 = 0\\\sqrt {z - 2021} - 2 = 0\end{array} \right.\)
\( \Leftrightarrow \left\{ \begin{array}{l}\sqrt {x - 2019} = 2\\\sqrt {y - 2020} = 2\\\sqrt {z - 2021} = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x - 2019 = 4\\y - 2020 = 4\\z - 2021 = 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 2023\,\,\,\left( {tm} \right)\\y = 2024\,\,\left( {tm} \right)\\z = 2025\,\,\,\left( {tm} \right)\end{array} \right..\)
Vậy phương trình có nghiệm duy nhất \(\left( {x;\,\,y;\,\,z} \right) = \left( {2023;\,\,2024;\,\,2025} \right).\)
Chọn D.
