Question

GPT : \(x^2+\sqrt{x+2006}=2006\)

Answer 1

Đặt \(\sqrt{x+2006}=a\left(a\ge0\right)\)

=> \(2006=a^2-x\)

Có \(x^2+a=a^2-x\)

<=> \(x^2-a^2+a-x=0\)

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

<=>\(\left(x-a\right)\left(x+a-1\right)=0\)

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

<=>\(\left[{}\begin{matrix}\sqrt{x+2006}=x\\x+\sqrt{x+2006}=1\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}x+2006=x^2\\\sqrt{x+2006}=1-x\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}x^2-x-2006=0\\x+2006=1-2x+x^2\end{matrix}\right.\)

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