Bài 12:

ĐKXĐ: $x \ge 0, x \ne 4, x \ne 9$.

a/ 

Đặt $A = \dfrac{x-3\sqrt{x}}{x-9} - 1$ và $B = \dfrac{9-x}{x+\sqrt{x}-6} - \dfrac{\sqrt{x}-3}{2-\sqrt{x}} - \dfrac{\sqrt{x}-2}{\sqrt{x}+3}$.

$A = \dfrac{\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)} - 1 = \dfrac{\sqrt{x}}{\sqrt{x}+3} - \dfrac{\sqrt{x}+3}{\sqrt{x}+3} = \dfrac{-3}{\sqrt{x}+3}$

$B = \dfrac{9-x}{(\sqrt{x}-2)(\sqrt{x}+3)} + \dfrac{\sqrt{x}-3}{\sqrt{x}-2} - \dfrac{\sqrt{x}-2}{\sqrt{x}+3}$

$B = \dfrac{9-x + (\sqrt{x}-3)(\sqrt{x}+3) - (\sqrt{x}-2)^2}{(\sqrt{x}-2)(\sqrt{x}+3)}$

$B = \dfrac{9-x + x - 9 - (x - 4\sqrt{x} + 4)}{(\sqrt{x}-2)(\sqrt{x}+3)} = \dfrac{-x + 4\sqrt{x} - 4}{(\sqrt{x}-2)(\sqrt{x}+3)}$

$B = \dfrac{-(\sqrt{x}-2)^2}{(\sqrt{x}-2)(\sqrt{x}+3)} = \dfrac{2-\sqrt{x}}{\sqrt{x}+3}$

$P = A : B = \dfrac{-3}{\sqrt{x}+3} \cdot \dfrac{\sqrt{x}+3}{2-\sqrt{x}} = \dfrac{3}{\sqrt{x}-2}$

Vậy $P = \dfrac{3}{\sqrt{x}-2}$.

b/ 

$P < 1 \Leftrightarrow \dfrac{3}{\sqrt{x}-2} - 1 < 0 \Leftrightarrow \dfrac{3 - \sqrt{x} + 2}{\sqrt{x}-2} < 0 \Leftrightarrow \dfrac{5 - \sqrt{x}}{\sqrt{x}-2} < 0$

TH1: $\begin{cases} 5 - \sqrt{x} > 0 \\ \sqrt{x} - 2 < 0 \end{cases} \Leftrightarrow \begin{cases} \sqrt{x} < 5 \\ \sqrt{x} < 2 \end{cases} \Leftrightarrow \sqrt{x} < 2 \Leftrightarrow x < 4$.

Kết hợp ĐKXĐ $\Rightarrow 0 \le x < 4$.

TH2: $\begin{cases} 5 - \sqrt{x} < 0 \\ \sqrt{x} - 2 > 0 \end{cases} \Leftrightarrow \begin{cases} \sqrt{x} > 5 \\ \sqrt{x} > 2 \end{cases} \Leftrightarrow \sqrt{x} > 5 \Leftrightarrow x > 25$.

Kết hợp ĐKXĐ $\Rightarrow x > 25$.

Vậy $0 \le x < 4$ hoặc $x > 25$.

Bài 13:

ĐKXĐ: $x \ge 0, x \ne 1$.

a/ 

$P = \dfrac{15\sqrt{x}-11}{(\sqrt{x}-1)(\sqrt{x}+3)} - \dfrac{3\sqrt{x}-2}{\sqrt{x}-1} - \dfrac{2\sqrt{x}+3}{\sqrt{x}+3}$

$P = \dfrac{15\sqrt{x}-11 - (3\sqrt{x}-2)(\sqrt{x}+3) - (2\sqrt{x}+3)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+3)}$

$P = \dfrac{15\sqrt{x}-11 - (3x + 7\sqrt{x} - 6) - (2x + \sqrt{x} - 3)}{(\sqrt{x}-1)(\sqrt{x}+3)}$

$P = \dfrac{15\sqrt{x} - 11 - 3x - 7\sqrt{x} + 6 - 2x - \sqrt{x} + 3}{(\sqrt{x}-1)(\sqrt{x}+3)}$

$P = \dfrac{-5x + 7\sqrt{x} - 2}{(\sqrt{x}-1)(\sqrt{x}+3)} = \dfrac{-(\sqrt{x}-1)(5\sqrt{x}-2)}{(\sqrt{x}-1)(\sqrt{x}+3)}$

Vậy $P = \dfrac{2-5\sqrt{x}}{\sqrt{x}+3}$.

b/ 

$P = \dfrac{1}{2} \Leftrightarrow \dfrac{2-5\sqrt{x}}{\sqrt{x}+3} = \dfrac{1}{2}$

$\Rightarrow 4 - 10\sqrt{x} = \sqrt{x} + 3 \Leftrightarrow 11\sqrt{x} = 1 \Leftrightarrow \sqrt{x} = \dfrac{1}{11} \Rightarrow x = \dfrac{1}{121}$ (TM ĐKXĐ).

Vậy $x = \dfrac{1}{121}$.

c/ 

Xét hiệu:

$P - \dfrac{2}{3} = \dfrac{2-5\sqrt{x}}{\sqrt{x}+3} - \dfrac{2}{3} = \dfrac{3(2-5\sqrt{x}) - 2(\sqrt{x}+3)}{3(\sqrt{x}+3)} = \dfrac{6 - 15\sqrt{x} - 2\sqrt{x} - 6}{3(\sqrt{x}+3)} = \dfrac{-17\sqrt{x}}{3(\sqrt{x}+3)}$

Với $x \ge 0 \Rightarrow -17\sqrt{x} \le 0$ và $3(\sqrt{x}+3) > 0$.

Do đó $\dfrac{-17\sqrt{x}}{3(\sqrt{x}+3)} \le 0 \Rightarrow P - \dfrac{2}{3} \le 0 \Rightarrow P \le \dfrac{2}{3}$ (đpcm).

Bài 14:

ĐKXĐ: $x \ge 0, x \ne m^2$.

a/ 

$P = \dfrac{2\sqrt{x}}{\sqrt{x}+m} + \dfrac{\sqrt{x}}{\sqrt{x}-m} - \dfrac{m^2}{4(x-m^2)}$

$P = \dfrac{4\cdot 2\sqrt{x}(\sqrt{x}-m) + 4\sqrt{x}(\sqrt{x}+m) - m^2}{4(\sqrt{x}-m)(\sqrt{x}+m)}$

$P = \dfrac{8x - 8m\sqrt{x} + 4x + 4m\sqrt{x} - m^2}{4(x-m^2)}$

$P = \dfrac{12x - 4m\sqrt{x} - m^2}{4(x-m^2)} = \dfrac{12x - 6m\sqrt{x} + 2m\sqrt{x} - m^2}{4(\sqrt{x}-m)(\sqrt{x}+m)}$

$P = \dfrac{6\sqrt{x}(2\sqrt{x}-m) + m(2\sqrt{x}-m)}{4(\sqrt{x}-m)(\sqrt{x}+m)}$

Vậy $P = \dfrac{(6\sqrt{x}+m)(2\sqrt{x}-m)}{4(\sqrt{x}-m)(\sqrt{x}+m)}$.

b/ 

$P = 0 \Rightarrow (6\sqrt{x}+m)(2\sqrt{x}-m) = 0$

Vì $m > 0$ và $x \ge 0 \Rightarrow 6\sqrt{x}+m > 0$.

Do đó: $2\sqrt{x}-m = 0 \Leftrightarrow 2\sqrt{x} = m \Leftrightarrow \sqrt{x} = \dfrac{m}{2} \Rightarrow x = \dfrac{m^2}{4}$.

Thử lại ĐKXĐ: $x = \dfrac{m^2}{4} \ne m^2$ (luôn đúng do $m > 0$).

Vậy $x = \dfrac{m^2}{4}$.

c/ 

Từ câu b, có $x = \dfrac{m^2}{4}$.

Yêu cầu bài toán $\Leftrightarrow x > 1 \Leftrightarrow \dfrac{m^2}{4} > 1 \Leftrightarrow m^2 > 4$.

Vì $m > 0 \Rightarrow m > 2$.

Vậy $m > 2$.