Question

giải phương trình

\(x^2+2x=2\sqrt{2x-1_{ }}\)

Answer 1

ĐK: x\(\ge\dfrac{1}{2}\)\(x^2+2x=2\sqrt{2x-1}\Leftrightarrow2x-2\sqrt{2x-1}+x^2=0\Leftrightarrow2x-1-2\sqrt{2x-1}+1+x^2=0\Leftrightarrow\left(\sqrt{2x-1}-1\right)^2+x^2=0\)(*)

Ta có \(\left(\sqrt{2x-1}-1\right)^2\ge0\) và \(x^2\ge0\)

Suy ra (*)\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{2x-1}-1\right)^2=0\\x^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2x-1}=1\\x=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x=2\\x=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)(ktm)

Vậy S=∅