Question

Giải phương trình sau: \(\sqrt{2023-\sqrt{x}}=2023-x\)

giúp mik vs ạ, mik cảm ơn :>

Answer 1

\(\sqrt{2023-\sqrt{x}}=2023-x\left(ĐK:x\ge0\right)\)

Đặt \(t=\sqrt{x}\left(t\le2023\right)\)

Pt trở thành : \(\sqrt{2023-t}=2023-t^2\)

\(\Leftrightarrow2023-t=\left(2023-t^2\right)^2\)

\(\Leftrightarrow t^4-4046t+4092529=2023-t\)

\(\Leftrightarrow t^4-4045+4090506=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2023\left(n\right)\\t=2022\left(n\right)\end{matrix}\right.\)

+) Với \(t=2023\Rightarrow x^2=2023\Rightarrow x=\pm17\sqrt{7}\)

+) Với \(t=2022\Rightarrow x^2=2022\Leftrightarrow x=\pm\sqrt{2022}\)

Vì \(x\ge0\) \(\Rightarrow x\in\left\{17\sqrt{7};\sqrt{2022}\right\}\)

Vậy \(S=\left\{17\sqrt{7};\sqrt{2022}\right\}\)