Đáp án:
$=(x^2-3x+1)(x^2-4x+1)$
Giải thích các bước giải:
$\begin{array}{l}x^4-7x^3+14x^2-7x+1\\=x^2(x^2-7x+14-\dfrac{7}{x}+\dfrac{1}{x^2})\\=x^2[(x^2+\dfrac{1}{x^2})-7(x+\dfrac{1}{x})+14]\\=x^2[(x^2+2+2+\dfrac{1}{x^2})-7(x+\dfrac{1}{x})+12]\\x^2[(x+\dfrac{1}{x})^2-7(x+\dfrac{1}{x})+12]\\=x^2[(x+\dfrac{1}{x})^2-3(x+\dfrac{1}{x})-4(x+\dfrac{1}{x})+12]\\=x^2[(x+\dfrac{1}{x})(x+\dfrac{1}{x}-3)-4(x+\dfrac{1}{x}-3)]\\=x^2(x+\dfrac{1}{x}-3)(x+\dfrac{1}{x}-4)\\=(x^2-3x+1)(x^2-4x+1)\\\end{array}$
