Bài 428
a) $A = \dfrac{x^2 - 4x + 1}{x^2}$
$A = 1 - \dfrac{4}{x} + \dfrac{1}{x^2}$
Đặt $t = \dfrac{1}{x}$.
$A = t^2 - 4t + 1 = (t - 2)^2 - 3$
$A_{min} = -3 \Leftrightarrow t - 2 = 0 \Leftrightarrow t = 2 \Leftrightarrow \dfrac{1}{x} = 2 \Leftrightarrow x = \dfrac{1}{2}$.

b) $B = \dfrac{4x^2 - 6x + 1}{(2x - 1)^2}$
ĐKXĐ: $x \ne \dfrac{1}{2}$.
$B = \dfrac{(2x - 1)^2 - 2x}{(2x - 1)^2} = 1 - \dfrac{2x}{(2x - 1)^2}$.
Xét $k = B$.
$k = \dfrac{4x^2 - 6x + 1}{(2x - 1)^2} \Leftrightarrow k(4x^2 - 4x + 1) = 4x^2 - 6x + 1$
$\Leftrightarrow (4k - 4)x^2 - (4k - 6)x + (k - 1) = 0$
$\Delta' = (2k - 3)^2 - (4k - 4)(k - 1) = 4k^2 - 12k + 9 - (4k^2 - 8k + 4) = -4k + 5$.
Để có nghiệm, $\Delta' \ge 0 \Leftrightarrow -4k + 5 \ge 0 \Leftrightarrow 4k \le 5 \Leftrightarrow k \le \dfrac{5}{4}$.
Vậy GTLN của $B$ là $\dfrac{5}{4}$ khi $x = -\dfrac{1}{2}$

429
a) $A = \dfrac{x}{(x + 10)^2}$
Nếu $x \le 0 \Rightarrow A \le 0$.
Nếu $x > 0$:
$A = \dfrac{x}{x^2 + 20x + 100} = \dfrac{1}{x + 20 + \dfrac{100}{x}}$
Để A max, mẫu min.
Theo BĐT Cauchy: $x + \dfrac{100}{x} \ge 2\sqrt{x \cdot \dfrac{100}{x}} = 20$.
$\Rightarrow x + 20 + \dfrac{100}{x} \ge 20 + 20 = 40$.
Vậy $A \le \dfrac{1}{40}$.
$A_{max} = \dfrac{1}{40} \Leftrightarrow x = \dfrac{100}{x} \Leftrightarrow x^2 = 100 \Leftrightarrow x = 10$ (do $x>0$).

b) 
$B = \dfrac{x}{(x + 100)^2}$
Nếu $x \le 0 \Rightarrow B \le 0$.
Nếu $x > 0$:
$B = \dfrac{x}{x^2 + 200x + 10000} = \dfrac{1}{x + 200 + \dfrac{10000}{x}}$
Để B max, mẫu min.
Theo BĐT Cauchy: $x + \dfrac{10000}{x} \ge 2\sqrt{x \cdot \dfrac{10000}{x}} = 200$.
$\Rightarrow x + 200 + \dfrac{10000}{x} \ge 200 + 200 = 400$.
Vậy $B \le \dfrac{1}{400}$.
$B_{max} = \dfrac{1}{400} \Leftrightarrow x = \dfrac{10000}{x} \Leftrightarrow x^2 = 10000 \Leftrightarrow x = 100$ (do $x>0$)