`a)`

`A = x  - 4 \sqrt{x} + 10 (x \ge 0)`

` = (x - 4 \sqrt{x} + 4) + 6`

`=  [ (\sqrt{x})^2 - 2 . \sqrt{x} . 2 + 2^2] + 6`

`=  (\sqrt{x} -2)^2 + 6`

`\forall x` ta có :

`(\sqrt{x} -2)^2 \ge 0`

`=> (\sqrt{x} - 2)^2+  6 \ge 6`

`=> A\ge 6`

Dấu `=` xảy ra `<=> \sqrt{x}  -2=0`

`<=>\sqrt{x}=2`

`=>x=4` (thỏa mãn điều kiện)

Vậy `\text{Min}_A = 6 <=> x= 4`

`b)`

`C =x - \sqrt{x} (x \ge 0)`

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

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

`=  (\sqrt{x} -1/2)^2 - 1/4`

`\forall x` ta có :

`(\sqrt{x} - 1/2)^2 \ge 0`

`=> (\sqrt{x} -1/2)^2 - 1/4 \ge -1/4`

`=> C \ge -1/4`

Dấu `=` xảy ra `<=> \sqrt{x} -1/2=0`

`<=>\sqrt{x}  =1/2`

`=>x=1/4` (thỏa mãn điều kiện)

Vậy `\text{Min}_C = -1/4 <=> x=1/4`

`c)`

`D = \sqrt{x^2-2x+4} +1`

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

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

`\forall x` ta có :

`(x-1)^2 \ge 0`

`=> (x-1)^2 + 3 \ge 3`

`=> \sqrt{ (x-1)^2 + 3} \ge \sqrt{3}`

`=> \sqrt{ (x-1)^2 + 3} + 1 \ge \sqrt{3} +1`

`=> D \ge \sqrt{3} +1`

Dấu `=` xảy ra `<=> (x-1)^2=0`

`<=>x-1=0`

`<=>x=1`

Vậy `\tex{Min}_D = 1 + \sqrt{3} <=>x=1`

``

`(\sqrt{\sqrt{11} + \sqrt{7} }  - \sqrt{\sqrt{11} - \sqrt{7} })^2`

`=  (\sqrt{ \sqrt{11} + \sqrt{7}})^2 - 2 . \sqrt{ \sqrt{11} + \sqrt{7}} . \sqrt{ \sqrt{11} - \sqrt{7} } + (\sqrt{ \sqrt{11} - \sqrt{7}})^2`

`=  (\sqrt{11} + \sqrt{7}) - 2 . \sqrt{ (\sqrt{11} + \sqrt{7}) . (\sqrt{11} - \sqrt{7}) } + (\sqrt{11}-  \sqrt{7})`

`=  \sqrt{11}+\sqrt{7}-2 . \sqrt{ (\sqrt{11})^2- (\sqrt{7})^2 } + \sqrt{11} - \sqrt{7}`

`=  \sqrt{11} + \sqrt{7}+ \sqrt{11} - \sqrt{7} - 2 \sqrt{ 11-7}`

` = 2 \sqrt{11}  - 2 . \sqrt{4}`

`=  2 \sqrt{11} - 2 . 2`

` = 2 \sqrt{11} -4`