Question

Giải phương trình :

a,\(x^2+\sqrt{x+2021}=2021\)

b,\(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)

Answer 1

a, \(x^2+\sqrt{x+2021}=2021\) ĐK \(x\ge-2021\)

<=> \(x^2-2021=-\sqrt{x+2021}\)

Đặt \(\sqrt{x+2021}=a\left(a\ge0\right)\)

=> \(\left\{{}\begin{matrix}x^2-2021=-a\\a^2-2021=x\end{matrix}\right.\)

=> \(\left(x-a\right)\left(x+a\right)+a+x=0\)

<=> \(\left[{}\begin{matrix}x+a=0\\x-a+1=0\end{matrix}\right.\)

+ \(x+a=0\)

=> \(\sqrt{x+2021}=-x\)

=> \(\left\{{}\begin{matrix}x\le0\\x^2-x-2021=0\end{matrix}\right.\)=> \(x=\frac{1-7\sqrt{165}}{2}\)

+ \(x-a+1=0\)

=> \(x+1=\sqrt{x+2021}\)

=> \(\left\{{}\begin{matrix}x\ge-1\\x^2+x-2020\end{matrix}\right.\)=> \(x=\frac{-1+\sqrt{8081}}{2}\)

Vậy \(S=\left\{\frac{-1+\sqrt{8081}}{2};\frac{1-7\sqrt{165}}{2}\right\}\)