(2x - 1)2 - 9 = 4\(\sqrt{x^2-x}\) Đk: x ≥ 1
⇔ (2x - 4)(2x + 2) = 4\(\sqrt{x^2-x}\)
⇔ 4(x - 2)(x + 1) =4 \(\sqrt{x^2-x}\)
⇔ (x - 2)(x + 1) = \(\sqrt{x^2-x}\)
⇔ x2 - x - 2 - \(\sqrt{x^2-x}\) = 0
⇔ (x2 - x) + 2\(\sqrt{x^2-x}\) + 1 - 3\(\sqrt{x^2-x}\) - 3 = 0
⇔ (\(\sqrt{x^2-x}\) + 1 )2 - 3(\(\sqrt{x^2-x}\) + 1) = 0
⇔ (\(\sqrt{x^2-x}\) + 1)(\(\sqrt{x^2-x}\) - 3) = 0
⇔ \(\left[{}\begin{matrix}\sqrt{x^2-x}=-1\\\sqrt{x^2-x}=3\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}\sqrt{x^2-x}=-1\\x^2-x=9\end{matrix}\right.\)(TH ở trên loại vì \(\sqrt{x^2-x}\)≥0
⇔ x2 - x - 9 = 0
⇔ (x - \(\frac{1}{2}\))2 - \(\frac{37}{4}\) = 0
⇔ (x - \(\frac{1+\sqrt{37}}{2}\))(x - \(\frac{1-\sqrt{37}}{2}\)) = 0
⇔ x = \(\frac{1\pm\sqrt{37}}{2}\)
Vậy nghiệm của pt là x = \(\frac{1\pm\sqrt{37}}{2}\)
PT \(\Leftrightarrow4x^2-4x-8-4\sqrt{x^2-x}=0\)
\(\Leftrightarrow4\left(x^2-x\right)-4\sqrt{x^2-x}+1=9\)
\(\Leftrightarrow\left(2\sqrt{x^2-x}-1\right)^2=9\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x^2-x}-1=3\\2\sqrt{x^2-x}-1=-3\end{matrix}\right.\)\(\Leftrightarrow2\sqrt{x^2-x}=4\)
\(\Leftrightarrow x^2-x-4=0\) (ĐKXĐ: \(x>1;x< 0\))
\(\Leftrightarrow x=\frac{1\pm\sqrt{17}}{2}\) (t/m)