Đáp án + Giải thích các bước giải:

 `K= (3\sqrt{x})/(\sqrt{x}-2)-(x+5\sqrt{x}-6)/(x-4)+(1-\sqrt{x})/(\sqrt{x}+2)`        Điều kiện: `x>=0; x\ne4`

`K= (3\sqrt{x})/(\sqrt{x}-2)-(x+5\sqrt{x}-6)/((\sqrt{x}-2)(\sqrt{x}+2))+(1-\sqrt{x})/(\sqrt{x}+2)` 

`K= (3\sqrt{x}(\sqrt{x}+2))/((\sqrt{x}-2)(\sqrt{x}+2))-(x+5\sqrt{x}-6)/((\sqrt{x}-2)(\sqrt{x}+2))+((1-\sqrt{x})(\sqrt{x}-2))/((\sqrt{x}-2)(\sqrt{x}+2))`

`K= (3\sqrt{x}(\sqrt{x}+2)-x+5\sqrt{x}-6+(1-\sqrt{x})(\sqrt{x}-2))/((\sqrt{x}-2)(\sqrt{x}+2))`

`K= (3x+6\sqrt{x}-x-5\sqrt{x}+6+\sqrt{x}-2-x+2\sqrt{x})/((\sqrt{x}-2)(\sqrt{x}+2))`

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

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

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

Vậy `K= (\sqrt{x}+2)/(\sqrt{x}-2)`.

`K=(3\sqrt{x})/(\sqrt{x}-2)-(x+5\sqrt{x}-6)/(x-4)+(1-\sqrt{x})/(\sqrt{x}+2)`

Điều kiện : `xne4` và `xge0`

`=(3\sqrt{x})/(\sqrt{x}-2)-(x+5\sqrt{x}-6)/((\sqrt{x}-2).(\sqrt{x}+2))+(1-\sqrt{x})/(\sqrt{x}+2)`

`=(3\sqrt{x}.(\sqrt{x}+2))/((\sqrt{x}-2).(\sqrt{x}+2))-(x+5\sqrt{x}-6)/((\sqrt{x}-2).(\sqrt{x}+2))-((\sqrt{x}-1).(\sqrt{x}-2))/((\sqrt{x}+2).(\sqrt{x}-2))`

`=((3x+6\sqrt{x}-x-5\sqrt{x)+6-x+3\sqrt{x}-2))/((\sqrt{x}-2).(\sqrt{x}+2))`

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

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

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

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