Question

[PHó]

rút gọn bthức sau

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

Answer 1

[Kiệt Nguyễn]

\(ĐKXĐ:x\ge0\)

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

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

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

\(=\frac{-2\sqrt{x}}{x-2\sqrt{x}}:\frac{8\sqrt{x}+4x}{4-x}\)

\(=\frac{-2\sqrt{x}}{x-2\sqrt{x}}.\frac{4-x}{8\sqrt{x}+4x}\)

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

\(=\frac{\left(2+\sqrt{x}\right)}{\sqrt{x}\left(4+2\sqrt{x}\right)}=\frac{1}{2\sqrt{x}}\)

Answer 2

[Inequalities]

mk ko kt lại nên sai từ dòng 2 r, bạn cộng thêm (3+căn x) vào r giải tương tự