Question

\(\left(\frac{x\sqrt{x}+1}{x-1}-\frac{x-1}{\sqrt{x}-1}\right):\left(\sqrt{x}+\frac{\sqrt{x}}{\sqrt{x}-1}\right)\)

Answer 1

Lời giải:

ĐKXĐ: \(x\neq 1; x\geq 0\)

\(P=\left[\frac{(\sqrt{x})^3+1}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}-1}\right]:\frac{x-\sqrt{x}+\sqrt{x}}{\sqrt{x}-1}\)

\(=\left[\frac{(\sqrt{x}+1)(x-\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}-(\sqrt{x}+1)\right].\frac{\sqrt{x}-1}{x}\)

\(=\left[\frac{x-\sqrt{x}+1}{\sqrt{x}-1}-(\sqrt{x}+1)\right].\frac{\sqrt{x}-1}{x}=\frac{x-\sqrt{x}+1-(\sqrt{x}+1)(\sqrt{x}-1)}{\sqrt{x}-1}.\frac{\sqrt{x}-1}{x}\)

\(=\frac{2-\sqrt{x}}{x}\)