\(2x^2+3x+3=5\sqrt{2x^2+3x+9}\)
\(\Leftrightarrow2x^2+3x+3=5\sqrt{2x^2+3x+3+6}\)(*)
Đặt \(2x^2+3x+3=a\)
(*) \(\Leftrightarrow a=5\sqrt{a+6}\)
\(\Leftrightarrow a^2=25\left(a+6\right)\)
\(\Leftrightarrow a^2-25a-150=0\)
\(\Leftrightarrow\left(a-30\right)\left(a+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=30\\a=-5\end{matrix}\right.\)
Trả lại biến cũ: \(2x^2+3x+3=30\Leftrightarrow2x^2+3x-27=0\)\(\Leftrightarrow\left(x-3\right)\left(2x+9\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{9}{2}\end{matrix}\right.\)
\(2x^2+3x+3=-5\Leftrightarrow2x^2+3x+8=0\)\(\Leftrightarrow\left(x\sqrt{2}+\frac{3\sqrt{2}}{4}\right)^2=-\frac{55}{8}\left(L\right)\)
Đat: \(2x^2+3x+3=a\)
\(\Rightarrow a=5\sqrt{a+6}\Leftrightarrow a^2=25a+150\Leftrightarrow a^2-25a-150=0\Leftrightarrow\left(a-12,5\right)^2=6,25\Leftrightarrow\left[{}\begin{matrix}a=10\\a=15\end{matrix}\right.\) \(+,a=10\Leftrightarrow x^2+3x+3=10\Leftrightarrow\left(x+\frac{3}{2}\right)^2=9,25\Leftrightarrow x=\pm\sqrt{9,25}-\frac{3}{2}\)
\(+,a=15\Leftrightarrow x^2+3x+2,25=14,25\Leftrightarrow\left(x+\frac{3}{2}\right)^2=14,25\Leftrightarrow x=\pm\sqrt{14,25}-\frac{3}{2}\)
