Lời giải:

a)

\(A=\left[\frac{4\sqrt{y}(\sqrt{y}-2)}{(2+\sqrt{y})(\sqrt{y}-2)}-\frac{8y}{(\sqrt{y}-2)(\sqrt{y}+2)}\right]:\left[\frac{\sqrt{y}-1}{\sqrt{y}(\sqrt{y}-2)}-\frac{2(\sqrt{y}-2)}{\sqrt{y}(\sqrt{y}-2)}\right]\)

\(=\frac{-4y-8\sqrt{y}}{(\sqrt{y}+2)(\sqrt{y}-2)}: \frac{3-\sqrt{y}}{\sqrt{y}(\sqrt{y}-2)}\)

\(=\frac{-4\sqrt{y}(\sqrt{y}+2)}{(\sqrt{y}+2)(\sqrt{y}-2)}.\frac{\sqrt{y}(\sqrt{y}-2)}{3-\sqrt{y}}\)

\(=\frac{4y}{\sqrt{y}-3}\)

b) Với \(y>0; y\neq 4; 9\):

Để \(A=2\Leftrightarrow \frac{4y}{\sqrt{y}-3}=2\Leftrightarrow 4y=2\sqrt{y}-6\)

\(\Leftrightarrow 4y-2\sqrt{y}=-6\)

\(\Leftrightarrow 3y+(\sqrt{y}-1)^2=-5< 0\) (vô lý với mọi $y>0$)

Do đó không tồn tại $y$ để $A=2$

ĐKXĐ: \(\hept{\begin{cases}y>0\\y\ne1\end{cases}}\)

a/ Ta có: \(A=\left[\frac{\sqrt{y}^3-1}{\sqrt{y}\left(\sqrt{y}-1\right)}-\frac{\sqrt{y}^3+1}{\sqrt{y}\left(\sqrt{y}+1\right)}\right]:\frac{2\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}\)

    \(=\left[\frac{\left(\sqrt{y}-1\right)\left(y+\sqrt{y}+1\right)}{\sqrt{y}\left(\sqrt{y}-1\right)}-\frac{\left(\sqrt{y}+1\right)\left(y-\sqrt{y}+1\right)}{\sqrt{y}\left(\sqrt{y}+1\right)}\right].\frac{\sqrt{y}+1}{2\left(\sqrt{y}-1\right)}\)

    \(=\left(\frac{y+\sqrt{y}+1-y+\sqrt{y}-1}{\sqrt{y}}\right).\frac{\sqrt{y}+1}{2\left(\sqrt{y}-1\right)}\)

       \(=\frac{2\sqrt{y}}{\sqrt{y}}.\frac{\sqrt{y}+1}{2\left(\sqrt{y}-1\right)}=\frac{\sqrt{y}+1}{\sqrt{y}-1}\)

b/ \(A=\frac{\sqrt{y}+1}{\sqrt{y}-1}=1+\frac{2}{\sqrt{y}-1}\)

    Để \(A\in Z\Rightarrow\left(\sqrt{y}-1\right)\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

   Với \(\sqrt{y}-1=1\Rightarrow\sqrt{y}=2\Rightarrow y=4\)

   Với \(\sqrt{y}-1=-1\Rightarrow\sqrt{y}=0\Rightarrow y=0\)(loại)

   Với \(\sqrt{y}-1=2\Rightarrow\sqrt{y}=3\Rightarrow y=9\)

  Với \(\sqrt{y}-1=-2\Rightarrow\sqrt{y}=-1\) (loại)

      Vậy y = 4 , y = 9

1,

\(A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\left(đk:a\ne0;1;2;a\ge0\right)\)

\(=\frac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{a^2-a}.\frac{a-2}{a+2}\)

\(=\frac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{a\left(a-1\right)}.\frac{a-2}{a+2}\)

\(=\frac{2a\left(a-1\right)\left(a-2\right)}{a\left(a-1\right)\left(a+2\right)}=\frac{2\left(a-2\right)}{a+2}\)

Để \(A=1\)\(=>\frac{2a-4}{a+2}=1< =>2a-4-a-2=0< =>a=6\)

2, 

a, Điều kiện xác định của phương trình là \(x\ne4;x\ge0\)

b, Ta có : \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)

\(=\frac{2\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}-\frac{\sqrt{x}-2}{x-4}\)

\(=\frac{2\sqrt{x}+2+2}{x-4}=\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{2}{\sqrt{x}-2}\)

c, Với \(x=3+2\sqrt{3}\)thì \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)

\(=\dfrac{xy\left(x^{\dfrac{1}{2}}+y^{\dfrac{1}{2}}\right)}{x^{\dfrac{1}{2}}+y^{\dfrac{1}{2}}}=xy\)

\(A=\dfrac{x^{\dfrac{3}{2}}y+xy^{\dfrac{3}{2}}}{\sqrt{x}+\sqrt{y}}=\left(x+y\right).\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\).

Sai đề