Đáp án + Giải thích các bước giải:

 `-` Bài 3:

`a)` Thay `x=9` vào `A` có:

$A = \frac{2(9) - 8\sqrt{9}}{\sqrt{9} + 5}$

$= \frac{18 - 8(3)}{3 + 5}$

$= \frac{18 - 24}{8}= \frac{-6}{8}$

$= -\frac{3}{4}$

`-` b)

$B = \left(\frac{2}{\sqrt{x} - 4} - \frac{5 - \sqrt{x}}{x - 16}\right) : \frac{\sqrt{x} + 1}{\sqrt{x} + 4}$

$ = \left(\frac{2}{\sqrt{x} - 4} - \frac{5 - \sqrt{x}}{(\sqrt{x} - 4)(\sqrt{x} + 4)}\right) : \frac{\sqrt{x} + 1}{\sqrt{x} + 4}$

$= \left(\frac{2(\sqrt{x} + 4) - (5 - \sqrt{x})}{(\sqrt{x} - 4)(\sqrt{x} + 4)}\right) : \frac{\sqrt{x} + 1}{\sqrt{x} + 4}$

$= \left(\frac{2\sqrt{x} + 8 - 5 + \sqrt{x}}{(\sqrt{x} - 4)(\sqrt{x} + 4)}\right) : \frac{\sqrt{x} + 1}{\sqrt{x} + 4}$

$ = \frac{3\sqrt{x} + 3}{(\sqrt{x} - 4)(\sqrt{x} + 4)} : \frac{\sqrt{x} + 1}{\sqrt{x} + 4}$

$ = \frac{3(\sqrt{x} + 1)}{(\sqrt{x} - 4)(\sqrt{x} + 4)} \cdot \frac{\sqrt{x} + 4}{\sqrt{x} + 1}$

$= \frac{3}{\sqrt{x} - 4}$

`-` c)

Có:

$P = A \cdot B = \frac{2x - 8\sqrt{x}}{\sqrt{x} + 5} \cdot \frac{3}{\sqrt{x} - 4}$

$= \frac{2\sqrt{x}(\sqrt{x} - 4)}{\sqrt{x} + 5} \cdot \frac{3}{\sqrt{x} - 4}$

$= \frac{6\sqrt{x}}{\sqrt{x} + 5}$

`+)` Thay `P` vào $\sqrt{2P - 1} = P - 2$:

$\sqrt{2\left(\frac{6\sqrt{x}}{\sqrt{x} + 5}\right) - 1} = \frac{6\sqrt{x}}{\sqrt{x} + 5} - 2$

`<=>`$\sqrt{\frac{12\sqrt{x} - (\sqrt{x} + 5)}{\sqrt{x} + 5}} = \frac{6\sqrt{x} - 2(\sqrt{x} + 5)}{\sqrt{x} + 5}$

`<=>`$\sqrt{\frac{11\sqrt{x} - 5}{\sqrt{x} + 5}} = \frac{4\sqrt{x} - 10}{\sqrt{x}+5}$

`<=>`$\frac{11\sqrt{x} - 5}{\sqrt{x} + 5} = \frac{(4\sqrt{x} - 10)^2}{(\sqrt{x} + 5)^2}$

`<=>`$(11\sqrt{x} - 5)(\sqrt{x} + 5) = (4\sqrt{x} - 10)^2$

`<=>`$11x + 55\sqrt{x} - 5\sqrt{x} - 25 = 16x - 80\sqrt{x} + 100$

`<=>`$11x + 50\sqrt{x} - 25 = 16x - 80\sqrt{x} + 100$

`<=>`$0 = 5x - 130\sqrt{x} + 125$

`<=>`$x - 26\sqrt{x} + 25 = 0$

`+)` Đặt `t=sqrt(x)` có:

`t^2 - 26t + 25 = 0`

`<=>(t - 1)(t - 25) = 0`

`<=>`\(\left[ \begin{array}{l}t=1\\t=25\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=1\\x=625\end{array} \right.\) 

`+)` Thế `x` vào PT ban đầu có:

`+)` Với `x=1` có:

$P = \frac{6\sqrt{1}}{\sqrt{1} + 5} = \frac{6}{6} = 1$ 

`=>`$\sqrt{2(1) - 1} = 1 - 2$

`=>`$ \sqrt{1} = -1$

`=>`$1 = -1 \text{ (vô lý)}$

`+)` Với `x=625` có:

$P = \frac{6\sqrt{625}}{\sqrt{625} + 5} = \frac{6(25)}{25 + 5} = \frac{150}{30} = 5$

`=>`$ \sqrt{2(5) - 1} = 5 - 2$

`=>`$\sqrt{9} = 3$

`=>`$ 3 = 3 \text{ (hợp lý)}$

`-` Bài 4:

`-` a)

$x = 4(\sqrt{9+4\sqrt{5}} - \sqrt{9-4\sqrt{5}})$

`+)` $9+4\sqrt{5} = (2+\sqrt{5})^2$

`+)` $9-4\sqrt{5} = (\sqrt{5}-2)^2$

`=>`$\sqrt{9+4\sqrt{5}} = 2+\sqrt{5}$

`=>`$\sqrt{9-4\sqrt{5}} = \sqrt{5}-2$

`<=>`$x = 4(2+\sqrt{5} - (\sqrt{5}-2)) = 4.4= 16$

`->`$B = \frac{\sqrt{16}-2}{\sqrt{16}+1}$

$= \frac{4-2}{4+1} = \frac{2}{5}$

`-` b)

$A = \frac{3(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)} + \frac{1}{\sqrt{x}-2} + \frac{\sqrt{x}-3}{\sqrt{x}}$

 $=\frac{3}{\sqrt{x}} + \frac{1}{\sqrt{x}-2} + \frac{\sqrt{x}-3}{\sqrt{x}}$

$= \frac{3(\sqrt{x}-2) + \sqrt{x} + (\sqrt{x}-3)(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)}$

$ = \frac{3\sqrt{x}-6 + \sqrt{x} + x - 2\sqrt{x} - 3\sqrt{x} + 6}{\sqrt{x}(\sqrt{x}-2)}$

$= \frac{x - \sqrt{x}}{\sqrt{x}(\sqrt{x}-2)}$

$= \frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-2)}$

$= \frac{\sqrt{x}-1}{\sqrt{x}-2}$

`-` c)

Có:

$A.B = \frac{\sqrt{x}-1}{\sqrt{x}-2} \cdot \frac{\sqrt{x}-2}{\sqrt{x}+1}$

$= \frac{\sqrt{x}-1}{\sqrt{x}+1}$

`<=>`$\sqrt{AB} = \sqrt{\frac{\sqrt{x}-1}{\sqrt{x}+1}} < \frac{2}{3}$

`<=>`$\frac{\sqrt{x}-1}{\sqrt{x}+1} < \frac{4}{9}$

`<=>`$9(\sqrt{x}-1) < 4(\sqrt{x}+1)$

`<=>`$9\sqrt{x} - 9 < 4\sqrt{x} + 4$

`<=>`$5\sqrt{x} < 13$

`<=>`$\sqrt{x} < \frac{13}{5}$

`<=>x < (13/5)^2 = 6,76`

`-` Vì `x>0`,`xne4` và `x` là số nguyên.

`=>`$x \in \{1, 2, 3, 5, 6\}$