Question

[hoàng thiên]

Rút gọn:

\(\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right):\frac{x-6\sqrt{x}+9}{\left(2-\sqrt{x}\right)\left(\sqrt{x}-3\right)}\)

Answer 1

[Duyên]

= \(\left[\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]:\frac{x-6\sqrt{x}+9}{\left(2-\sqrt{x}\right)\left(\sqrt{x}-3\right)}\)

= \(\left[\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}+\frac{4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right]:\frac{\left(\sqrt{x}-3\right)^2}{\left(2-\sqrt{x}\right)\left(\sqrt{x}-3\right)}\)

= \(\left[\frac{\left(2+\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}-\frac{\left(2-\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right]:\frac{\sqrt{x}-3}{2-\sqrt{x}}\)

= \(\left[\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{nt}\right]:nt\)

\(=\left[\frac{8\sqrt{x}+4x}{nt}\right]:nt\)

\(=\left[\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right]:nt\)

\(=\frac{4\sqrt{x}}{2-\sqrt{x}}.\frac{\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)

\(=\frac{4\sqrt{x}}{\sqrt{x}-3}\)

Answer 2

[💋Amanda💋]

https://i.imgur.com/N39fVia.jpg