Question

giải pt: \(x^2+\sqrt{x+2021}=2021\)

Answer 1

ĐKXĐ \(x\ge-2021\)

Đặt \(\sqrt{x+2021}=a\ge0\Rightarrow a^2=x+2021\Rightarrow2021=a^2-x\)

phương trình trở thành : 

\(x^2+a=a^2-x\Leftrightarrow\left(x-a\right)\left(x+a\right)+a+x=0\Leftrightarrow\left(x+a\right)\left(x-a+1\right)=0\)

TH1: x+a=0

 \(\Leftrightarrow x+\sqrt{x+2021}=0\Leftrightarrow x+2021=x^2\left(x\le0\right)\\ \Leftrightarrow x=\dfrac{1-7\sqrt{165}}{2}\left(t.m\right)\)

TH2: x+1=a

\(\Leftrightarrow x+1=\sqrt{x+2021}\Leftrightarrow x^2+2x+1=x+2021\left(x\ge-1\right)\\ \Leftrightarrow x=\dfrac{-1+\sqrt{8081}}{2}\left(t.m\right)\)