ĐKXĐ `x > 0 ; x \ne 4`

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

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

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

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

`B = 7/3A = 14/( 3\sqrt{x} + 6 )`

`Do  x > 0 => B = 14/( 3\sqrt{x} + 6 ) > 0  ( 1 )`

`B - 3 = ( 14 - 9\sqrt{x} - 18 )/( 3\sqrt{x} + 6 ) = ( -9\sqrt{x} - 4 )/( 3\sqrt{x} + 6 ) < 0  AA  x > 0`

`B < 3  ( 2 )`

`Mà  B in Z  ( 3 )`

`( 1 ) ; ( 2 ) ; ( 3 ) => B in { 1 ; 2 }`

`+B = 1 => 14 = 3\sqrt{x} + 6`

`3\sqrt{x} = 8`

`\sqrt{x} = 8/3`

`x = 64/9` ( TMĐKXĐ )

`+B = 2 => 7 = 3\sqrt{x} + 6`

`3\sqrt{x} = 1`

`\sqrt{x} = 1/3`

`x = 1/9` ( TMĐKXĐ )

`Vậy  x in { 64/9 ; 1/9 }` thì `B in Z`