a) ĐK : \(x\ne1\)
\(A=\left(\frac{x+\sqrt{x}+1}{x+\sqrt{x}-2}-\frac{1}{1+\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{x-1}\)
\(=(\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{x+\sqrt{x}+1+\sqrt{x}+2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)})\) \(\cdot\left(x-1\right)\)
\(=(\frac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)})\cdot\left(x-1\right)=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\left(x-1\right)=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}=\)
\(\left(\sqrt{x}+1\right)^2\)
\(\frac{1}{A}\) là một số tự nhiên
* Với \(\sqrt{x}\in I\Rightarrow\frac{1}{A}\in I\) => Không thỏa mãn
* Với \(\sqrt{x}\in Z\Rightarrow x\in Z\Rightarrow\left(\sqrt{x}+1\right)^2\in Z\)
\(\Rightarrow\frac{1}{\left(\sqrt{x}+1\right)^2}\in Z\) khi
\(\sqrt{x}+1\inƯ\left(1\right)=\left\{\pm1\right\}\)
=> x = 0 ( thỏa mãn ) hoặc \(\sqrt{x}=-2\) ( vô lí )
Vậy x = 0 là giá trị cần tìm
ĐKXĐ: \(x\ge0;x\ne1\)
\(A=\left(\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}\right)\left(x-1\right)\)
\(A=\left(\frac{x+\sqrt{x}+1+\sqrt{x}+2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(A=\left(\frac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(A=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\left(\sqrt{x}+1\right)^2\)
Do \(A=\left(\sqrt{x}+1\right)^2\ge1\Rightarrow\frac{1}{A}\le1\)
\(\Rightarrow\frac{1}{A}\) là số tự nhiên khi và chỉ khi \(A=1\Leftrightarrow\left(\sqrt{x}+1\right)=1\Rightarrow x=0\)
