ĐKXĐ: x>=0; x<>1
\(B=\left(\dfrac{2x+1}{\sqrt{x^3}-1}-\dfrac{\sqrt[]{x}}{x+\sqrt{x}+1}\right)\cdot\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\left(\dfrac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\cdot\left(\dfrac{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\dfrac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(1-\sqrt{x}+x-\sqrt{x}\right)\)
\(=\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2\)
\(=\dfrac{x+\sqrt{x}+1}{\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)=\sqrt{x}-1\)
Lời giải:
ĐKXĐ: $x\geq 0; x\neq 1$
\(B=\left[\frac{2x+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\right].\left[\frac{(\sqrt{x}+1)(x-\sqrt{x}+1)}{\sqrt{x}+1}-\sqrt{x}\right]\)
\(=\frac{2x+1-\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}(x-\sqrt{x}+1-\sqrt{x})\\ =\frac{x+\sqrt{x}+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}(x-2\sqrt{x}+1)\\ =\frac{1}{\sqrt{x}-1}(\sqrt{x}-1)^2=\sqrt{x}-1\)
