`1)` ĐKXĐ `x >= 0 ; x \ne 1 ; 4`

`a) Q = (( x - 2\sqrt{x} )/( x - 4 ) - ( x - x\sqrt{x} - 6 )/( x + \sqrt{x} - 2 ) + ( \sqrt{x} + 1 )/( 1 - \sqrt{x} )) . ( x + 39 )/( x + 3\sqrt{x} - 10 )`

`Q = ((\sqrt{x}(\sqrt{x}-2))/((\sqrt{x}-2)(\sqrt{x}+2))+(x\sqrt{x}-x+6)/((\sqrt{x}-1)(\sqrt{x}+2))-((\sqrt{x}+1)(\sqrt{x}+2))/((\sqrt{x}-1)(\sqrt{x}+2))) . ( x + 39 )/(( \sqrt{x} - 2 )( \sqrt{x} + 5 ))`

`Q = ((\sqrt{x})/(\sqrt{x}+2) + ( x\sqrt{x} - x + 6 )/(( \sqrt{x} - 1 )( \sqrt{x} +2 )) - ( x + 3\sqrt{x} + 2 )/(( \sqrt{x}-  1 )( \sqrt{x} + 2 ))) . ( x + 39 )/(( \sqrt{x} - 2 )( \sqrt{x} + 5 ))`

`Q = ((\sqrt{x}(\sqrt{x}-1))/((\sqrt{x}-1)(\sqrt{x}+2)) + ( x\sqrt{x} - 2x - 3\sqrt{x} + 4 )/(( \sqrt{x} - 1 )( \sqrt{x} + 2 ))) . ( x + 39 )/(( \sqrt{x} - 2 )( \sqrt{x} + 5 ))`

`Q = ( x - \sqrt{x} + x\sqrt{x} - 2x - 3\sqrt{x} + 4 )/(( \sqrt{x} - 1 )( \sqrt{x} + 2 )) . ( x + 39 )/(( \sqrt{x} - 2 )( \sqrt{x} + 5 ))`

`Q = ( x\sqrt{x} - x - 4\sqrt{x} + 4 )/(( \sqrt{x} - 1 )( \sqrt{x} + 2 )) . ( x + 39 )/(( \sqrt{x} - 2 )( \sqrt{x} + 5 ))`

`Q = ( x( \sqrt{x} - 1 ) - 4( \sqrt{x} - 1 ))/(( \sqrt{x} - 1 )( x - 4 )( \sqrt{x} + 5 )) . ( x + 39 )`

`Q = (( \sqrt{x} - 1 )( x - 4 ))/(( \sqrt{x} - 1 )( x - 4 )) . ( x + 39 )/( \sqrt{x} + 5 )`

`Q = ( x + 39 )/( \sqrt{x} + 5 )`

`Vậy  Q = ( x + 39 )/( \sqrt{x} + 5 )` với `x >= 0 ; x \ne 1 ; 4`

`b) Q = ( x - 25 + 64 )/( \sqrt{x} + 5 )`

`Q = \sqrt{x} - 5 + 64/( \sqrt{x} + 5 )`

`Q = \sqrt{x} + 5 + 64/( \sqrt{x} + 5 ) -10`

Áp dụng BĐT AM - GM có:

`\sqrt{x} + 5 + 64/( \sqrt{x} + 5 ) >= 2\sqrt{( \sqrt{x} + 5 ) . 64/( \sqrt{x} + 5 )} = 16`

`Q = \sqrt{x} + 5 + 64/( \sqrt{x} + 5 ) - 10 >= 6`

`Dấu = xảy   ra:  ( \sqrt{x} + 5 )^2 = 64`

`\sqrt{x} + 5 = 8  ( do  \sqrt{x} + 5 > 0  AA  x >= 0 )`

`\sqrt{x} = 3`

`x = 9` ( TMĐKXĐ )

`Vậy  Q min = 6` khi `x = 9`