`(1/(\sqrt{a}-1)-1/\sqrt{a}):((\sqrt{a}+1)/(\sqrt{a}-2)-(\sqrt{a}+2)/(\sqrt{a}-1))`

ĐKXĐ : `a>0,ane4,ane1`

`=(\sqrt{a}/(\sqrt{a}.(\sqrt{a}-1))-(\sqrt{a}-1)/(\sqrt{a}.(\sqrt{a}-1))):(((\sqrt{a}+1).(\sqrt{a}-1))/((\sqrt{a}-2).(\sqrt{a}-1))-((\sqrt{a}+2).(\sqrt{a}-2))/((\sqrt{a}-2).(\sqrt{a}-1)))`

`=(\sqrt{a}-\sqrt{a}+1)/(\sqrt{a}.(\sqrt{a}-1)):(((\sqrt{a})^2-1^2)/((\sqrt{a}-2).(\sqrt{a}-1))-((\sqrt{a})^2-2^2)/((\sqrt{a}-2).(\sqrt{a}-1)))`

`=(1)/(\sqrt{a}.(\sqrt{a}-1)):((a-1)/((\sqrt{a}-2).(\sqrt{a}-1))-(a-4)/((\sqrt{a}-2).(\sqrt{a}-1)))`

`=(1)/(\sqrt{a}.(\sqrt{a}-1)):(a-1-a+4)/((\sqrt{a}-2).(\sqrt{a}-1))`

`=(1)/(\sqrt{a}.(\sqrt{a}-1)).((\sqrt{a}-2).(\sqrt{a}-1))/3`

`=(\sqrt{a}-2)/(3\sqrt{a})`

`b)`

Để `A>1/6`

`<=>(\sqrt{a}-2)/(3\sqrt{a})>1/6`

`<=>(\sqrt{a}-2)/(3\sqrt{a})-1/6>0`

`<=>(6.(\sqrt{a}-2))/(18\sqrt{a})-(3\sqrt{a})/(18\sqrt{a})>0`

`<=>(6\sqrt{a}-12)/(18\sqrt{a})-(3\sqrt{a})/(18\sqrt{a})>0`

`<=>(3\sqrt{a}-12)/(18\sqrt{a})>0`

Trường hợp `1:`

`{(3\sqrt{a}-12>0),(),(18\sqrt{a}>0):}`

`<=>{(3\sqrt{a}>12),(),(\sqrt{a}>0):}`

`<=>{(\sqrt{a}>4),(),(\sqrt{a}>0):}`

`<=>{(a>16),(),(a>0):}`

`<=>a>16`

Trường hợp `2:`

`{(3\sqrt{a}-12<0),(),(18\sqrt{a}<0):}`

`<=>{(3\sqrt{a<12),(),(\sqrt{a}<0):}`

`<=>{(\sqrt{a}<4),(),(\sqrt{a}<0):}`

`<=>{(a<16),(),(a<0):}`

`<=>a<0` ( không thỏa mãn ĐKXĐ )

Vậy `a>16` thì `A>1/6`