Question

Rút gọn \(\left(\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{8\sqrt{x}+8}{x+2\sqrt{x}}-\frac{\sqrt{x}+2}{\sqrt{x}}\right):\frac{x+2\sqrt{x}+5}{x+2\sqrt{x}}\)

Answer 1

đkxđ: x>0

\(\left(\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{8\sqrt{x}+8}{x+2\sqrt{x}}-\frac{\sqrt[]{x}+2}{\sqrt{x}}\right):\frac{x+2\sqrt{x}+5}{x+2\sqrt{x}}\)

= \(\left[\frac{\sqrt{x}.\sqrt{x}+8\sqrt{x}+8-\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\sqrt{x}}\right]:\frac{x+2\sqrt{x}+5}{x+2\sqrt{x}}\)

=\(\frac{4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\sqrt{x}}.\frac{\left(\sqrt{x}+2\right)\sqrt{x}}{x+2\sqrt{x}+5}\)

=\(\frac{4\sqrt{x}+4}{x+2\sqrt{x}+5}\)