ĐKXĐ `x >= 0 ; x \ne 9`

`a) C = ( \sqrt{x} )/( \sqrt{x} + 3 ) + ( 2\sqrt{x} )/(\sqrt{x} - 3)  + ( 3x + 9 )/( 9 - x )`

`C = (\sqrt{x}(\sqrt{x}-3))/((\sqrt{x}-3)(\sqrt{x}+3))+(2\sqrt{x}(\sqrt{x}+3))/((\sqrt{x}-3)(\sqrt{x}+3))-(3x+9)/((\sqrt{x}-3)(\sqrt{x}+3))`

`C = ( x - 3\sqrt{x} + 2x + 6\sqrt{x} - 3x - 9 )/(( \sqrt{x} - 3 )( \sqrt{x} + 3 ))`

`C = ( 3\sqrt{x} + 9 )/(( \sqrt{x} - 3 )( \sqrt{x} + 3 ))`

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

`C = 3/( \sqrt{x} - 3 )`

`Vậy  C = 3/( \sqrt{x} - 3 )` với `x >= 0 ;x \ne 9`

`b) | x - 5 | = 2`

`x - 5 = 2  hoặc  x - 5 = -2`

`x = 7  hoặc   x = 3` ( TMĐKXĐ )

`+) x = 7` thay vào `C` có:

`C = 3/( \sqrt{7} - 3 ) = ( 3( \sqrt{7} + 3))/( 7 - 9 ) = ( 3\sqrt{7} + 9 )/( -2 ) = (-3\sqrt{7}-9)/2`

`+) x = 3` thay vào `C` có:

`C = 3/(\sqrt{3}-3) = ( 3(\sqrt{3}+3))/( 3 - 9 ) = ( 3(\sqrt{3}+3))/(-6) = (-\sqrt{3}-3)/2`

`Vậy  x = 3` thì `C = ( -\sqrt{3} - 3 )/2 ; x = 7` thì `C = ( -3\sqrt{7} - 9 )/2`

`c) C = 3/( \sqrt{x} - 3 )`

`Do  x >= 0 => \sqrt{x} >= 0`

`\sqrt{x} - 3 >= -3`

`C = 3/( \sqrt{x} - 3 ) <= -1`

`dấu = xảy  ra:  x = 0`

`Vậy  C max = -1` khi `x = 0`