\(3x-2\sqrt{4x-3}=3\) (ĐK: \(x\ge1\))
\(\Leftrightarrow2\sqrt{4x-3}=3x-3\)
\(\Leftrightarrow\left(2\sqrt{4-3}\right)^2=\left(3x-3\right)^2\)
\(\Leftrightarrow4\cdot\left(4x-3\right)=9x^2-18+9\)
\(\Leftrightarrow16x-12-9x^2+18x-9=0\)
\(\Leftrightarrow34x-9x^2-21=0\)
\(\Leftrightarrow27x+7x-9x^2-21=0\)
\(\Leftrightarrow\left(27x-9x^2\right)-\left(21-7x\right)=0\)
\(\Leftrightarrow9x\left(3-x\right)-7\left(3-x\right)=0\)
\(\Leftrightarrow\left(3-x\right)\left(9x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x=0\\9x-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(n\right)\\x=\frac{7}{9}\left(l\right)\end{matrix}\right.\)
Vậy: x=3
