a: \(B=\dfrac{x+\sqrt{x}-1-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{1}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
b: \(B-\dfrac{1}{3}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{3}=\dfrac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{-\left(x-2\sqrt{x}+1\right)}{3\left(x+\sqrt{x}+1\right)}=\dfrac{-\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\)
Do đó: B<1/3
c: \(B=\dfrac{1}{2\sqrt{x}+1}\)
\(\Leftrightarrow2x+\sqrt{x}=x+\sqrt{x}+1\)
=>x=1(loại)
\(a,\\ =\dfrac{x\sqrt{x}+x-\sqrt{x}+x+\sqrt{x}-1-x+1}{\left(\sqrt{x+1}\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{x\sqrt{x}+x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\left(\sqrt{x}-1\right)\)
\(\dfrac{x\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{x}{x+\sqrt{x}+1}\\ b,\dfrac{x}{x+\sqrt{x}+1}=\dfrac{x}{x+2\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}\\ =\dfrac{x}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2}< \dfrac{1}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}=\dfrac{1}{3}\)
\(c,\Leftrightarrow\dfrac{1}{x+\sqrt{x}+1}=\dfrac{1}{2\sqrt{x}+1}\\ \Leftrightarrow x+\sqrt{x}+1=2\sqrt{x}+1\\ \Leftrightarrow x-\sqrt{x}=0\\ \Rightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\\ \Rightarrow\left\{{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
