Question

giải phương trình: x² - 2x = 2√2x-1

Answer 1

\(x^2-2x=2\sqrt{2x-1}\) \(\left(Đk:x\ge\dfrac{1}{2}\right)\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2x-1}+1\\x=-\sqrt{2x-1}-1\end{matrix}\right.\)

+) \(x=\sqrt{2x-1}+1\)

\(x-1=\sqrt{2x-1}\left(x\ge1\right)\)

\(x^2-2x+1=2x-1\)

\(x^2-4x+2=0\)

\(\left(x-2\right)^2=2\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}+2\left(TM\right)\\x=2-\sqrt{2}\left(L\right)\end{matrix}\right.\)

+) \(x=-\sqrt{2x-1}-1\)

VP\(\le-1\) mà \(VT\ge\dfrac{1}{2}\)

=> phương trình vô nghiệm

Vậy \(S=\left\{2+\sqrt{2}\right\}\)