\(x^4-3x^3+4x^2-3x+1=0\\ \Leftrightarrow x^2\left(x^2-3x+4-\dfrac{3}{x}+\dfrac{1}{x^2}\right)=0\\ \Leftrightarrow x^2\left[\left(x^2+2+\dfrac{1}{x^2}\right)-\left(3x+\dfrac{3}{x}\right)+2\right]=0\\ \Leftrightarrow x^2\left[\left(x+\dfrac{1}{x}\right)^2-3\left(x+\dfrac{1}{x}\right)+2\right]=0\\ Đặt\text{ }x+\dfrac{1}{x}=y\\ \Leftrightarrow x^2\left(y^2-3y+2\right)=0\\ \Leftrightarrow x^2\left(y^2-2y-y+2\right)=0\\ \Leftrightarrow x^2\left[\left(y^2-2y\right)-\left(y-2\right)\right]=0\\ \Leftrightarrow x^2\left[y\left(y-2\right)-\left(y-2\right)\right]=0\\ \Leftrightarrow x^2\left(y-1\right)\left(y-2\right)=0\\ \Leftrightarrow x^2\left(x+\dfrac{1}{x}-1\right)\left(x+\dfrac{1}{x}-2\right)=0\\ \Leftrightarrow\left(x^2+1-x\right)\left(x^2+1-2x\right)=0\\ \Leftrightarrow\left(x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\right)\left(x-1\right)^2=0\\ \Leftrightarrow\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)^2=0\left(Vì\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ne0\right)\\ \Leftrightarrow x-1=0\\ \Leftrightarrow x=1\\Vậy\text{ }S=\left\{1\right\}\)
