a/ Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\)
\(\Leftrightarrow2\left(x^2+\frac{1}{x^2}\right)-3\left(x-\frac{1}{x}\right)-4=0\)
Đặt \(x-\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2+2\)
Pt trở thành:
\(2\left(t^2+2\right)-3t-4=0\Leftrightarrow2t^2-3t=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=\frac{3}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{x}=0\\x-\frac{1}{x}=\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\2x^2-3x-2=0\end{matrix}\right.\) (bấm máy)
b/
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=3\)
\(\Leftrightarrow\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-3\right)-3=0\)
\(\Leftrightarrow\left(x^2-5x+4\right)\left(x^2-5x+6\right)-3=0\)
Đặt \(x^2-5x+4=t\)
Pt trở thành:
\(t\left(t+2\right)-3=0\)
\(\Leftrightarrow t^2+2t-3=0\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x+4=1\\x^2-5x+4=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x+3=0\\x^2-5x+7=0\end{matrix}\right.\) (bấm máy)
