Đkxđ: `x geq 0 ; x !=4`

`A = 1/(\sqrt{x} + 2) + 1/(\sqrt{x} - 2) + x/(x - 4)`

`A = (\sqrt{x} - 2 + \sqrt{x} + 2 + x)/(x - 4)`

`A = (x + 2\sqrt{x})/((\sqrt{x} - 2)(\sqrt{x} + 2))`

`A = (\sqrt{x}(\sqrt{x} + 2))/((\sqrt{x} - 2)(\sqrt{x} + 2))`

`A = \sqrt{x}/(\sqrt{x} - 2)`

`a. A < 1 <=> \sqrt{x}/(\sqrt{x} - 2) < 1`

`\iff \sqrt{x}/(\sqrt{x} - 2) - 1 < 0`

`\iff (\sqrt{x} - (\sqrt{x} - 2))/(\sqrt{x} - 2) < 0`

`\iff 2/(\sqrt{x} - 2) < 0`

Do `2>0` nên `\sqrt{x} - 2 < 0 \iff \sqrt{x} < 2 \iff x < 4`

`to 0 leqx <4`

`b. -19 + 8\sqrt{3} = 16 + 2 \cdot 4 \cdot \sqrt{3} + 3 = (4 + \sqrt{3})^2`

`-19 - 8\sqrt{3} = 16 - 2 \cdot 4 \cdot \sqrt{3} + 3 = (4 - \sqrt{3})^2`

`x = 1/2 (\sqrt{(4 + \sqrt{3})^2} + \sqrt{(4 - \sqrt{3})^2}) - 1`

`x = 1/2 (4 + \sqrt{3} + 4 - \sqrt{3}) - 1`

`x = 1/2 \cdot 8 - 1 = 3` (t/m)

`B = (\sqrt{x} + 3)/(2 - \sqrt{x}) : (2A)`

`B = -(\sqrt{x} + 3)/(\sqrt{x} - 2) : (2\sqrt{x})/(\sqrt{x} - 2)`

`B = -(\sqrt{x} + 3)/(\sqrt{x} - 2) \cdot (\sqrt{x} - 2)/(2\sqrt{x})`

`B = -(\sqrt{x} + 3)/(2\sqrt{x})`

Thay `x=3` vào `B` ta có:

`B = -(\sqrt{3} + 3)/(2\sqrt{3}) = -(\sqrt{3}(1 + \sqrt{3}))/(2\sqrt{3}) = -(1 + \sqrt{3})/2`

`c.` Đkxđ: `0 leq x leq 9 ; x ne4`

Thay `A = \sqrt{x}/(\sqrt{x} - 2)` vào vế trái (VT) của phương trình:

`VT = \sqrt{x}/(\sqrt{x} - 2) \cdot (\sqrt{x} - 2) + 5\sqrt{x} = \sqrt{x} + 5\sqrt{x} = 6\sqrt{x}= x + 4 + \sqrt{x + 16} + \sqrt{9 - x}`

`\iff (x - 6\sqrt{x} + 9) + (\sqrt{x + 16} - 5) + \sqrt{9 - x} = 0`

`\iff (\sqrt{x} - 3)^2 + (\sqrt{x + 16} - 5) + \sqrt{9 - x} = 0`

`\sqrt{x + 16} - 5 = (x + 16 - 25)/(\sqrt{x + 16} + 5) = (x - 9)/(\sqrt{x + 16} + 5)`

`to(\sqrt{x} - 3)^2 + (x - 9)/(\sqrt{x + 16} + 5) + \sqrt{9 - x} = 0`

`-` Xét `(\sqrt{x} - 3)^2 >= 0`

Với `0 leq x leq 9 \implies \sqrt{x} leq 3`

`-> \sqrt{x} - 3 leq 0` và `\sqrt{x} + 3 > 0`

`-> ((\sqrt{x} - 3)(\sqrt{x} + 3))/(\sqrt{x + 16} + 5) <= 0`

`\sqrt{9 - x} geq 0`

PT đã cho: `6\sqrt{x} = x + 4 + \sqrt{x + 16} + \sqrt{9 - x}`

`(x - 6\sqrt{x} + 9) + (\sqrt{x + 16} + \sqrt{9 - x} - 5) = 0`

`\iff (\sqrt{x} - 3)^2 + (\sqrt{x + 16} + \sqrt{9 - x} - 5) = 0`

`-(\sqrt{x} - 3)^2 >= 0`

Dấu "=" xảy ra `\iff \sqrt{x} = 3 \iff x = 9`

`-` `\sqrt{x + 16} + \sqrt{9 - x} - 5`

`->(\sqrt{x + 16} + \sqrt{9 - x})^2 = (x + 16) + (9 - x) + 2\sqrt{(x + 16)(9 - x)}`

`= 25 + 2\sqrt{(x + 16)(9 - x)}`

Vì `0 <= x <= 9` 

`-> x + 16 > 0` hoặc `9 - x >= 0`

`-> (\sqrt{x + 16} + \sqrt{9 - x})^2 >= 25`

`->\sqrt{x + 16} + \sqrt{9 - x} >= 5`

`-> \sqrt{x + 16} + \sqrt{9 - x} - 5 >= 0`

Dấu "=" xảy ra `\iff (x + 16)(9 - x) = 0 \iff x = -16` (loại) hoặc `x=9` (t/m)

Vậy `x=9` t/m y/c bài toán