Câu hỏi của RIBFUBUG

Question

giải phương trình

$x^2+2018x-2017=2\sqrt{2020x-2019}$

Nguyễn Việt Lâm · Accepted Answer

ĐKXĐ:...

$\Leftrightarrow x^2-2x+1+2020x-2019-2\sqrt{2020x-2019}+1=0$

$\Leftrightarrow\left(x-1\right)^2+\left(\sqrt{2020x-2019}-1\right)^2=0$

$\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\sqrt{2020x-2019}-1=0\end{matrix}\right.$

$\Rightarrow x=1$

Câu trả lời:

ĐKXĐ:...

$\Leftrightarrow x^2-2x+1+2020x-2019-2\sqrt{2020x-2019}+1=0$

$\Leftrightarrow\left(x-1\right)^2+\left(\sqrt{2020x-2019}-1\right)^2=0$

$\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\sqrt{2020x-2019}-1=0\end{matrix}\right.$

$\Rightarrow x=1$