\(A=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)

vì x nguyên để A nhận GT nguyên thì \(\frac{2}{\sqrt{x}+1}\in Z\)\(\Rightarrow2⋮\sqrt{x}+1\)\(\Rightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

Mà \(\sqrt{x}+1\ge0+1=1\)nên \(\sqrt{x}+1\in\left\{1;2\right\}\Rightarrow\sqrt{x}\in\left\{0;1\right\}\)

\(\Rightarrow x\in\left\{0;1\right\}\)

Đây là bài của lớp 7 mà

Ta có \(A=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)

Vì \(x\in Z\Rightarrow\sqrt{x}\in Z\)

Để \(A\in Z\Rightarrow\frac{2}{\sqrt{x}+1}\in Z\)

\(\Rightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)               

Mà  \(\sqrt{x}\ge0\) \(\forall x\)

\(\Rightarrow\sqrt{x}+1\ge1\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x}+1=1\\\sqrt{x}+1=2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=1;x=-1\end{cases}}\)

Vậy \(x\in\left\{0;1;-1\right\}\)

ĐKXĐ: \(x.y\ge1\)

Áp dụng BĐT AM - GM: \(x\sqrt{y-1}+y\sqrt{x-1}=x\sqrt{1\left(y-1\right)}+y\sqrt{1\left(x-1\right)}\)

\(\le x.\frac{1+y-1}{2}+y.\frac{1+x-1}{2}=xy\)

Đẳng thức xảy ra khi \(\hept{\begin{cases}1=y-1\\1=x-1\end{cases}}\Rightarrow x=y=2\)

Thay xuống pt dưới thấy thỏa mãn.

Vậy x = y = 2

P/s: Em chưa học về hệ pt nên ko chắc ạ, giải đại.

\(ĐKXĐ:x\ne1;x\ne0\)

\(A=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{2\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{2x-2\sqrt{x}}{2x+2\sqrt{x}}\)

\(N=\frac{\sqrt{x}-3}{2\sqrt{x}}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)2\sqrt{x}}=\frac{x-2\sqrt{x}-3}{2x+2\sqrt{x}}\)

Ta có :

 \(x\ge0>-3\)

\(\Leftrightarrow x>-3\)

\(\Leftrightarrow x+\left(x-2\sqrt{x}\right)>-3+\left(x-2\sqrt{x}\right)\)

\(\Leftrightarrow2x-2\sqrt{x}>x-2\sqrt{x}-3\)

\(\Leftrightarrow\frac{2x-2\sqrt{x}}{2x+2\sqrt{x}}>\frac{x-2\sqrt{x}-3}{2x+2\sqrt{x}}\)

\(\Leftrightarrow A>N\)