\(x^2+6x-3=4x\sqrt{2x-1}\left(1\right)\) ĐK: \(x\ge\frac{1}{2}\)
Đặt \(\sqrt{2x-1}=a\ge0\)
\(\Rightarrow6x-3=3a^2\)
=> (1) <=> x^2 +3a^2 = 4ax
<=> x^2 -4ax +3a^2 =0
<=> x^2 -ax - 3ax + 3a^2 =0
<=> x(x-a) -3a(x-a) =0
<=> (x-a) ( x-3a ) =0
\(\Leftrightarrow\orbr{\begin{cases}x=a\\x=3a\end{cases}}\)
TH1: x=a
\(\Rightarrow x=\sqrt{2x-1}\)\(\left(x\ge0\right)\)
\(\Leftrightarrow x^2=2x-1\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
<=> x=1 (tm)
TH2: x= 3a
\(\Rightarrow x=3\sqrt{2x-1}\left(x\ge0\right)\)
\(\Leftrightarrow x^2=18x-9\)
\(\Leftrightarrow x^2-18x+9=0\)
\(\Delta=288\)
=> pt có 2 nghiệm pb \(\orbr{\begin{cases}x=\frac{18+12\sqrt{2}}{2}=9+6\sqrt{2}\left(tm\right)\\x=\frac{18-12\sqrt{2}}{2}=9-6\sqrt{2}\left(tm\right)\end{cases}}\)
Vậy ...
