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
