\(A=\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}+\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\left(x>0,x\ne1\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x+1}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}+\dfrac{x-\sqrt{x}+1}{\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}=\dfrac{3x+3}{\sqrt{x}}\)
\(A-\left|A\right|=0\Rightarrow A=\left|A\right|\Rightarrow\left[{}\begin{matrix}A=A\\A=-A\end{matrix}\right.\)
mà \(x>0\Rightarrow A>0\Rightarrow A=A\) (luôn đúng với mọi \(x\in R\) )
Ta có: \(A=\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}+\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}+1+x-\sqrt{x}+1+x+1}{\sqrt{x}}\)
\(=\dfrac{3x+3}{\sqrt{x}}\)
Để A-|A|=0 thì A=|A|
\(\Leftrightarrow3x+3\ge0\)
hay \(x\ge-1\)
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
