a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)
\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2-3\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x-2\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b: Khi \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\) thì \(A=\dfrac{2\cdot\sqrt{\left(\sqrt{3}-1\right)^2}-1}{\sqrt{\left(\sqrt{3}-1\right)^2}+1}\)
\(=\dfrac{2\left(\sqrt{3}-1\right)-1}{\sqrt{3}-1+1}\)
\(=\dfrac{2\sqrt{3}-2-1}{\sqrt{3}}=\dfrac{2\sqrt{3}-3}{\sqrt{3}}=2-\sqrt{3}\)
c: A=-0,5=-1/2
=>\(\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=-\dfrac{1}{2}\)
=>\(2\left(2\sqrt{x}-1\right)=-\sqrt{x}-1\)
=>\(4\sqrt{x}-2=-\sqrt{x}-1\)
=>\(4\sqrt{x}+\sqrt{x}=-1+2\)
=>\(5\sqrt{x}=1\)
=>\(\sqrt{x}=\dfrac{1}{5}\)
=>x=1/25(nhận)
d: Để A>0 thì \(\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}>0\)
=>\(2\sqrt{x}-1>0\)
=>\(2\sqrt{x}>1\)
=>\(\sqrt{x}>\dfrac{1}{2}\)
=>x>1/4
Để A<0 thì \(\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}< 0\)
=>\(2\sqrt{x}-1< 0\)
=>\(2\sqrt{x}< 1\)
=>\(\sqrt{x}< \dfrac{1}{2}\)
=>0<=x<1/4
e: Để A nguyên âm thì \(\left\{{}\begin{matrix}A< 0\\2\sqrt{x}-1⋮\sqrt{x}+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}0< =x< \dfrac{1}{4}\\2\sqrt{x}+2-3⋮\sqrt{x}+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}0< =x< \dfrac{1}{4}\\-3⋮\sqrt{x}+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0< =x< \dfrac{1}{4}\\\sqrt{x}+1\in\left\{1;3\right\}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}0< =x< \dfrac{1}{4}\\\sqrt{x}\in\left\{0;2\right\}\end{matrix}\right.\Leftrightarrow x=0\)
f: \(A\left(\sqrt{x}+1\right)=-\sqrt{x}\)
=>\(\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\left(\sqrt{x}+1\right)=-\sqrt{x}\)
=>\(2\sqrt{x}-1=-\sqrt{x}\)
=>\(3\sqrt{x}=1\)
=>\(\sqrt{x}=\dfrac{1}{3}\)
=>x=1/9(nhận)
