Question

TÌM ĐKXĐ VÀ RÚT GỌN BIỂU THỨC:   \(P=\left(\frac{1}{\sqrt{x}+2}+\frac{\sqrt{x}}{x-4}\right):\frac{2}{\sqrt{x}-2}\)

Answer 1

ĐKXĐ: \(\hept{\begin{cases}\sqrt{x}\ne2\\x\ne4\end{cases}\Rightarrow x\ne4}\)

\(P=\left[\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right).\left(\sqrt{x+2}\right)}\right]:\frac{2.\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x+2}\right)}\)

\(P=\frac{2\sqrt{x}-2}{\left(\sqrt{x}-2\right).\left(\sqrt{x+2}\right)}\cdot\frac{\left(\sqrt{x}-2\right).\left(\sqrt{x+2}\right)}{2\sqrt{x}+4}=\frac{2\sqrt{x}-2}{2\sqrt{x}+4}\)