a) Tìm giá trị nhỏ nhất của Q = x^2-4x+1/x^2 .

b) Ta có:

$$\begin{gathered} \left(a-\frac{1}{b}\right)\left(b-\frac{1}{c}\right)\left(c-\frac{1}{a}\right) \\ =\left(\frac{ab-1}{b}\right)\left(\frac{bc-1}{c}\right)\left(\frac{ac-1}{a}\right) \\ =\frac{\left(ab-1\right)\left(bc-1\right)\left(ac-1\right)}{a.b.c} \\ =\frac{\left(a{b}^{2}c-ab-bc+1\right)\left(ac-1\right)}{a.b.c} \\ =\frac{a^2{b}^2{c}^2-a{b}^{2}c-a^{2}bc+ab-ab{c}^2+bc+ac+1}{a.b.c} \\ =abc-a-b-c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a.b.c} \\ =abc-(a+b+c)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{1}{a.b.c} \end{gathered}$$

Để $\left(a-\frac{1}{b}\right)\left(b-\frac{1}{c}\right)\left(c-\frac{1}{a}\right)\in \mathbb{N}*$  thì a, b, c Î $\mathrm{\mathbb{N}}*$ và a.b.c Î Ư(1) = {1}.

Với a = b = c = 1

Ta thay vào:

$$\begin{gathered} abc-(a+b+c)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{1}{a.b.c} \\ =\mathrm{1.1.1}-(1+1+1)+(1+1+1)+1=2 \end{gathered}$$

Vậy a = b = c = 1.