Question

Giải phương trình \(x^2+2017x-2016=2\sqrt{2019x-2018}\)

Answer 1

\(DK:x\ge\frac{2018}{2019}\)

\(PT\Leftrightarrow x^2-2x+1+2019x-2018-2\sqrt{2019x-2018}+1=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(\sqrt{2019x-2018}-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(\sqrt{2019x-2018}-1\right)^2=0\end{cases}}\Leftrightarrow x=1\left(TM\right)\)