Đáp án:

`a)A={2a+2\sqrt{a}+2}/{\sqrt{a}}` 

`b)a\in{4;1/4}`

Giải thích các bước giải:

`ĐKXĐ:{(a>0),(a\ge1):}`

`a)A={a\sqrt{a}-1}/{a-\sqrt{a}}-{a\sqrt{a}+1}/{a+\sqrt{a}}+(\sqrt{a}-{1}/{\sqrt{a}})({\sqrt{a}+1}/{\sqrt{a}-1}+{\sqrt{a}-1}/{\sqrt{a}+1})`

`={(\sqrt{a}-1)(a+\sqrt{a}+1)}/{\sqrt{a}(\sqrt{a}-1)}-{(\sqrt{a}+1)(a-\sqrt{a}+1)}/{\sqrt{a}(\sqrt{a}+1)}+({a}/{\sqrt{a}}-{1}/{\sqrt{a}})[{(\sqrt{a}+1)(\sqrt{a}+1)}/{(\sqrt{a}+1)(\sqrt{a}-1)}+{(\sqrt{a}-1)(\sqrt{a}-1)}/{(\sqrt{a}+1)(\sqrt{a}-1)}]`

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

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

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

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

`={a+\sqrt{a}+1}/{\sqrt{a}}-{a-\sqrt{a}+1}/{\sqrt{a}}+{2a+2}/{\sqrt{a}}`

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

`={a+\sqrt{a}+1-a+\sqrt{a}-1+2a+2}/{\sqrt{a}}`

`={2a+2\sqrt{a}+2}/{\sqrt{a}}`

Vậy với `a>0,a\ge1 `thì `A={2a+2\sqrt{a}+2}/{\sqrt{a}}`

`b)A=7`

`⇔{2a+2\sqrt{a}+2}/{\sqrt{a}}=7`

`⇔{2a+2\sqrt{a}+2}/{\sqrt{a}}={7\sqrt{a}}/{\sqrt{a}}`

`⇒2a+2\sqrt{a}+2=7\sqrt{a}`

`⇔2a+2\sqrt{a}+2-7\sqrt{a}=0`

`⇔2a-5\sqrt{a}+2=0`

`⇔2a-4\sqrt{a}-\sqrt{a}+2=0`

`⇔2\sqrt{a}(\sqrt{a}-2)-(\sqrt{a}-2)=0`

`⇔(\sqrt{a}-2)(2\sqrt{a}-1)=0`

⇔ \(\left[ \begin{array}{l}\sqrt{a}-2=0\\2\sqrt{a}-1=0\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}\sqrt{a}=2\\2\sqrt{a}=1\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}\sqrt{a}=2\\\sqrt{a}=\frac{1}{2}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}a=4\\a=\frac{1}{4}\end{array} \right.\) 

Vậy với `a\in{4;1/4}` thì `A=7`