$b)$ $x^4 - 3x^3 - 6x^2 + 3x + 1$

$= x^2 \left(x^2 - 3x - 6 + \frac{3}{x} + \frac{1}{x^2}\right)$

$= x^2 \left[\left(x^2 + \frac{1}{x^2}\right) - 3\left(x - \frac{1}{x}\right) - 6\right]$

$= x^2 \left[\left(x - \frac{1}{x}\right)^2 + 2 - 3\left(x - \frac{1}{x}\right) - 6\right]$

$= x^2 \left[\left(x - \frac{1}{x}\right)^2 - 3\left(x - \frac{1}{x}\right) - 4\right]$

* Đặt $t = x - \frac{1}{x}$, ta được:

$t^2 - 3t - 4$

$= t^2 - 4t + t - 4$

$= t(t - 4) + 1(t - 4)$

$= (t - 4)(t + 1)$ $(1)$

* Thay $t = x - \frac{1}{x}$ vào $(1)$ ta được:

$= x^2 \left(x - \frac{1}{x} - 4\right)\left(x - \frac{1}{x} + 1\right)$

$= \left[x \left(x - \frac{1}{x} - 4\right)\right] \cdot \left[x \left(x - \frac{1}{x} + 1\right)\right]$

$= (x^2 - 4x - 1)(x^2 + x - 1)$

 

$d)$ $x^4 + 6x^3 - 11x^2 + 6x + 1$

$= x^2 \left(x^2 + 6x - 11 + \frac{6}{x} + \frac{1}{x^2}\right)$

$= x^2 \left[\left(x^2 + \frac{1}{x^2}\right) + 6\left(x + \frac{1}{x}\right) - 11\right]$

$= x^2 \left[\left(x + \frac{1}{x}\right)^2 - 2 + 6\left(x + \frac{1}{x}\right) - 11\right]$

$= x^2 \left[\left(x + \frac{1}{x}\right)^2 + 6\left(x + \frac{1}{x}\right) - 13\right]$

* Đặt $t = x + \frac{1}{x}$, ta được:

$t^2 + 6t - 13$

$= x^2 \left[\left(x + \frac{1}{x}\right)^2 + 6\left(x + \frac{1}{x}\right) - 13\right]$

$= x^2 \left(x^2 + \frac{1}{x^2} + 2 + 6x + \frac{6}{x} - 13\right)$

$= x^4 + 6x^3 - 11x^2 + 6x + 1$

 

$f)$ $x^4 - 7x^3 + 14x^2 - 7x + 1$

$= x^2 \left(x^2 - 7x + 14 - \frac{7}{x} + \frac{1}{x^2}\right)$

$= x^2 \left[\left(x^2 + \frac{1}{x^2}\right) - 7\left(x + \frac{1}{x}\right) + 14\right]$

$= x^2 \left[\left(x + \frac{1}{x}\right)^2 - 2 - 7\left(x + \frac{1}{x}\right) + 14\right]$

$= x^2 \left[\left(x + \frac{1}{x}\right)^2 - 7\left(x + \frac{1}{x}\right) + 12\right]$

* Đặt $t = x + \frac{1}{x}$, ta được:

$t^2 - 7t + 12$

$= t^2 - 3t - 4t + 12$

$= t(t - 3) - 4(t - 3)$

$= (t - 3)(t - 4)$ $(1)$

* Thay $t = x + \frac{1}{x}$ vào $(1)$ ta được:

$= x^2 \left(x + \frac{1}{x} - 3\right)\left(x + \frac{1}{x} - 4\right)$

$= \left[x \left(x + \frac{1}{x} - 3\right)\right] \cdot \left[x \left(x + \frac{1}{x} - 4\right)\right]$

$= (x^2 - 3x + 1)(x^2 - 4x + 1)$

$\color{#006600}{k}\color{#008800}{i}\color{#00AA00}{l}\color{#00CC00}{l}\color{#00EE00}{u}\color{#00FF00}{a}\color{#33FF33}{r}\color{#55FF55}{2}\color{#22FF22}{0}\color{#00FF00}{5}$