a/
\(\left(2x-1\right)\left(3x-1\right)\left(x-2\right)\left(x-3\right)=4x^2\)
\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)\left(3x-1\right)\left(x-3\right)=4x^2\)
\(\Leftrightarrow\left(2x^2-5x+2\right)\left(3x^2-10x+3\right)=4x^2\)
\(\Leftrightarrow\left(6x^2-15x+6\right)\left(6x^2-20x+6\right)=24x^2\)
Nhận thấy \(x=0\) không phải nghiệm, chia 2 vế cho \(x^2\):
\(\left(6x+\frac{6}{x}-15\right)\left(6x+\frac{6}{x}-20\right)=24\)
Đặt \(6x+\frac{6}{x}-20=a\Rightarrow6x+\frac{6}{x}-15=a+5\)
\(\left(a+5\right)a-24=0\Leftrightarrow a^2+5a-24=0\) \(\Rightarrow\left[{}\begin{matrix}a=3\\a=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}6x+\frac{6}{x}-20=3\\6x+\frac{6}{x}-20=-8\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}6x^2-23x+6=0\\6x^2-12x+6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{23\pm\sqrt{385}}{12}\\x=1\end{matrix}\right.\)
b/
\(3x^2-10x+6-\sqrt{2\left(x^4+4x^2+4-4x^2\right)}=0\)
\(\Leftrightarrow x^2-6x+2+2\left(x^2-2x+2\right)-\sqrt{2\left(x^2-2x+2\right)\left(x^2+2x+2\right)}=0\)
\(\Leftrightarrow x^2-6x+2+\sqrt{2\left(x^2-2x+2\right)}\left(\sqrt{2\left(x^2-2x+2\right)}-\sqrt{x^2+2x+2}\right)=0\)
\(\Leftrightarrow x^2-6x+2+\sqrt{2\left(x^2-2x+2\right)}\left(\frac{x^2-6x+2}{\sqrt{2\left(x^2-2x+2\right)}+\sqrt{x^2+2x+2}}\right)=0\)
\(\Leftrightarrow\left(x^2-6x+2\right)\left(1+\frac{\sqrt{2\left(x^2-2x+2\right)}}{\sqrt{2\left(x^2-2x+2\right)}+\sqrt{x^2+2x+2}}\right)=0\)
\(\Leftrightarrow x^2-6x+2=0\) (ngoặc to phía sau luôn dương)
\(\Rightarrow\left[{}\begin{matrix}x=3+\sqrt{7}\\x=3-\sqrt{7}\end{matrix}\right.\)
