`a)` `A= 1+({2a+\sqrt{a}-1}/{1-a}-{2a\sqrt{a}-\sqrt{a}+a}/{1-a\sqrt{a}}).{a-\sqrt{a}}/{2\sqrt{a}-1}`
$ĐKXĐ: \begin{cases}a\ge 0\\1-a\ne 0\\1-a\sqrt{a}\ne 0\\2\sqrt{a}-1\ne 0\end{cases}$`<=>`$ \begin{cases}a\ge 0\\a\ne 1\\a\ne \dfrac{1}{4}\end{cases}$
`=>a\ge 0;a\ne 1; a\ne 1/ 4`
$\\$
`A=1+({2a+\sqrt{a}-1}/{(1-\sqrt{a})(1+\sqrt{a})}-{\sqrt{a}(2a+\sqrt{a}-1)}/{1^3-(\sqrt{a})^3}).{a-\sqrt{a}}/{2\sqrt{a}-1}`
`=1+(2a+\sqrt{a}-1)(1/{(1-\sqrt{a})(1+\sqrt{a})}-\sqrt{a}/{(1-\sqrt{a})(a+\sqrt{a}+1)}).{\sqrt{a}(\sqrt{a}-1)}/{2\sqrt{a}-1}`
`=1+(2\sqrt{a}-1)(\sqrt{a}+1).{a+\sqrt{a}+1-\sqrt{a}(1+\sqrt{a})}/{(1-\sqrt{a})(1+\sqrt{a}).(a+\sqrt{a}+1)}.{-\sqrt{a}(1-\sqrt{a})}/{2\sqrt{a}-1}`
`=1+(2\sqrt{a}-1). 1/{a+\sqrt{a}+1} . {-\sqrt{a}}/{2\sqrt{a}-1}`
`=1-\sqrt{a}/{a+\sqrt{a}+1}`
`={a+\sqrt{a}+1-\sqrt{a}}/{a+\sqrt{a}+1}={a+1}/{a+\sqrt{a}+1}`
Vậy: `A={a+1}/{a+\sqrt{a}+1}` với `a\ge 0;a\ne 1;a\ne 1/4`
$\\$
`b)` `A=\sqrt{6}/{1+\sqrt{6}}`
`<=>{a+1}/{a+\sqrt{a}+1}=\sqrt{6}/{1+\sqrt{6}}`
`<=>(1+\sqrt{6})(a+1)=\sqrt{6}(a+\sqrt{a}+1)`
`<=>1.(a+1)+\sqrt{6}.(a+1)=\sqrt{6}.\sqrt{a}+\sqrt{6}.(a+1)`
`<=>a+1-\sqrt{6}.\sqrt{a}=0`
`<=>a-\sqrt{6}.\sqrt{a}+1=0` `(1)`
Đặt `t=\sqrt{a}\quad (t\ge 0; t\ne 1; 1/ 2)`
`(1)<=>t^2-\sqrt{6}t+1=0`
`<=>`$\left[\begin{array}{l}t=\dfrac{\sqrt{6}+\sqrt{2}}{2}\\t=\dfrac{\sqrt{6}-\sqrt{2}}{2}\end{array}\right.$
+) `TH: t={\sqrt{6}+\sqrt{2}}/2`
`=>a=t^2={6+2\sqrt{12}+2}/4={8+4\sqrt{3}}/4`
`=2+\sqrt{3}` (thỏa mãn)
$\\$
+) `TH: t={\sqrt{6}-\sqrt{2}}/2`
`=>a=t^2={6-2\sqrt{12}+2}/4={8-4\sqrt{3}}/4`
`=2-\sqrt{3}` (thỏa mãn)
Vậy `a\in {2-\sqrt{3};2+\sqrt{3}}` thỏa mãn đề bài
