a) \(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{x-1}{x+\sqrt{x}+1}\right)=\left[\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right].\dfrac{x+\sqrt{x}+1}{x-1}=\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{x-1}=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(x-1\right)}=\dfrac{1}{x-1}\)
b) Khi x=5+2\(\sqrt{3}\Leftrightarrow P=\dfrac{1}{5+2\sqrt{3}-1}=\dfrac{1}{4+2\sqrt{3}}=\dfrac{4-2\sqrt{3}}{\left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right)}=\dfrac{4-2\sqrt{3}}{16-12}=\dfrac{4-2\sqrt{3}}{4}=\dfrac{2\left(2-\sqrt{3}\right)}{4}=\dfrac{2-\sqrt{3}}{2}\)
c) Ta có \(\left|A\right|\le1\Leftrightarrow\left|\dfrac{1}{x-1}\right|\le1\Leftrightarrow\dfrac{1}{\left|x-1\right|}\le1\Leftrightarrow\left|x-1\right|\ge1\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-1\ge1\\1-x\le1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x\ge2\\x\le0\end{matrix}\right.\)
Kết hợp với ĐK
Vậy x\(\le0\) hoặc \(x\ge2\) thì \(\left|A\right|\le1\)
