\(\text{b) }6x^4+25x^3+12x^2-25x+6=0\\ \Leftrightarrow x^2\left(6x^2+25x+12-\dfrac{25}{x}+\dfrac{6}{x^2}\right)=0\\ \Leftrightarrow x^2\left(6x^2-12+\dfrac{6}{x^2}+25x-\dfrac{25}{x}+24\right)=0\\ \Leftrightarrow x^2\left[\left(6x^2-12+\dfrac{6}{x^2}\right)+\left(25x-\dfrac{25}{x}\right)+24\right]=0\\ \Leftrightarrow x^2\left[6\left(x^2-2+\dfrac{1}{x^2}\right)+25\left(x-\dfrac{1}{x}\right)+24\right]=0\\ \Leftrightarrow x^2\left[6\left(x-\dfrac{1}{x}\right)^2+25\left(x-\dfrac{1}{x}\right)+24\right]=0\)

Đặt \(x-\dfrac{1}{x}=t\)

\(\Leftrightarrow x^2\left(6t^2+25t+24\right)=0\\ \Leftrightarrow x^2\left(6t^2+9t+16t+24\right)=0\\ \Leftrightarrow x^2\left[3t\left(2t+3\right)+8\left(2t+3\right)\right]=0\\ \Leftrightarrow x^2\left(3t+8\right)\left(2t+3\right)=0\\ \Leftrightarrow x^2\left(3x-\dfrac{3}{x}+8\right)\left(2x-\dfrac{2}{x}+3\right)=0\\ \Leftrightarrow\left(3x^2-3+8x\right)\left(2x^2-2+3x\right)=0\\ \Leftrightarrow\left(3x^2+9x-x-3\right)\left(2x^2+4x-x-2\right)=0\\ \Leftrightarrow\left[3x\left(x+3\right)-\left(x+3\right)\right]\left[2x\left(x+2\right)-\left(x+2\right)\right]=0\\ \Leftrightarrow\left(3x-1\right)\left(x+3\right)\left(2x-1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x+3=0\\2x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\x=-3\\2x=1\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-3\\x=\dfrac{1}{2}\\x=-2\end{matrix}\right.\)

Vậy tập nghiệm phương trình là \(S=\left\{-\dfrac{1}{3};-3;\dfrac{1}{2};-2\right\}\)