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Trả lời:
a, \(B=\left(\frac{x+\sqrt{x}-1}{x\sqrt{x}-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\right):\frac{1}{\sqrt{x}-1}\left(ĐK:x>0;x\ne1\right)\)
\(=\left(\frac{x+\sqrt{x}-1}{\left(\sqrt{x}\right)^3-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\right):\frac{1}{\sqrt{x}-1}\)
\(=\left(\frac{x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\right).\left(\sqrt{x}-1\right)\)
\(=\left(\frac{x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\left(\sqrt{x}-1\right)\)
\(=\frac{x+\sqrt{x}-1-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\frac{x+\sqrt{x}-1-\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\frac{x+\sqrt{x}-1-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
b, \(B< \frac{1}{3}\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}< \frac{1}{3}\)
\(\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{3}< 0\)
\(\Leftrightarrow\frac{3\sqrt{x}}{3\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{3\left(x+\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{-x+2\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{-\left(x-2\sqrt{x}+1\right)}{3\left(x+\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{-\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\)
Vì \(-\left(\sqrt{x}-1\right)^2< 0\) với mọi \(x>0;x\ne1\)
\(3\left(x+\sqrt{x}+1\right)>0\) với mọi \(x>0;x\ne1\)
\(\Rightarrow\frac{-\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\) luôn đúng với mọi \(x>0;x\ne1\)
Vậy \(B< \frac{1}{3}\)
c, \(B=\frac{1}{2\sqrt{x}+1}\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}=\frac{1}{2\sqrt{x}+1}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)=x+\sqrt{x}+1\)
\(\Leftrightarrow2x+\sqrt{x}=x+\sqrt{x}+1\)
\(\Leftrightarrow x=1\) (tm)
Vậy x = 1 là giá trị cần tìm
\(Q=\left(\frac{\sqrt{x}^2-1}{2\sqrt{x}}\right)^2.\left[\frac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(Q=\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{2\sqrt{x}}\right].\left[\frac{\left(\sqrt{x}-1+\sqrt{x}+1\right)\left(\sqrt{x}-1-\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(Q=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{2\sqrt{x}}.\frac{-4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(Q=\frac{-4\sqrt{x}}{2\sqrt{x}}=-2\)