a)\(B=\left(\frac{2x+1}{\sqrt{x}^3-1}-\frac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\frac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)(\(ĐKXĐ:\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\))
\(=\left(\frac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(\frac{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\left(\frac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(1-\sqrt{x}+x-\sqrt{x}\right)\)
\(=\left(\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(x-2\sqrt{x}+1\right)\)
\(=\frac{1}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)^2=\sqrt{x}-1\)
b)Để B=3 thì \(\sqrt{x}-1=3\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)
