\(2x^2-6x-1=\sqrt{4x+5}\)
Đk:\(x\ge-\frac{5}{4}\)
\(\Leftrightarrow\left(2x^2-6x-1\right)^2=4x+5\)
\(\Leftrightarrow4x^4-23x^3+32x^2+12x+1=4x+5\)
\(\Leftrightarrow4x^4-24x^3+32x^2+8x-4=0\)
\(\Leftrightarrow4\left(x^4-6x^3+8x^2+2x-1\right)=0\)
\(\Leftrightarrow\left(x^2-4x+1\right)\left(x^2-2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-4x+1=0\left(1\right)\\x^2-2x-1=0\left(2\right)\end{cases}}\)
\(\Delta_{\left(1\right)}=\left(-2\right)^2-\left(-4\left(1.1\right)\right)=8\)
\(\Leftrightarrow x=\frac{2-\sqrt{8}}{2}\left(tm\right)\)
\(\Delta_{\left(2\right)}=\left(-4\right)^2-4\left(1.1\right)=12\)
\(\Leftrightarrow x=\frac{4+\sqrt{12}}{2}\left(tm\right)\)
