Tìm x, y biết (x - 1)^2022 + (căn bậc hai (y - 2))^2023 = 0

(x – 1)²⁰²² + \({\left( {\sqrt {y - 2} } \right)^{2023}} = 0\)

Do: \(\left\{ \begin{array}{l}{\left( {x - 1} \right)^{2022}} = {\left[ {{{\left( {x - 1} \right)}^{1011}}} \right]^2} \ge 0\\\sqrt {y - 2} \ge 0 \Leftrightarrow {\left( {\sqrt {y - 2} } \right)^{2023}} = 0\end{array} \right.\)

⇔ (x – 1)²⁰²² + \({\left( {\sqrt {y - 2} } \right)^{2023}} \ge 0\)

Khi (x – 1)²⁰²² + \({\left( {\sqrt {y - 2} } \right)^{2023}} = 0\)

Thì: \(\left\{ \begin{array}{l}x - 1 = 0\\\sqrt {y - 2} = 0\end{array} \right.\)

⇔\(\left\{ \begin{array}{l}x = 1\\y = 2\end{array} \right.\)

Vậy x = 1; y = 2.