Question

RÚT GỌN BIỂU THỨC SAU:

26) \(A = \left(\dfrac{\sqrt{x}}{\sqrt{x} + 2} - \dfrac{3}{2 - \sqrt{x}} + \dfrac{3\sqrt{x} - 2}{x - 2}\right) : \left(\dfrac{\sqrt{x} + 3}{\sqrt{x} - 2} + \dfrac{2\sqrt{x}}{2\sqrt{x} - x}\right)\)

 

Answer 1

Sửa đề: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{3}{2-\sqrt{x}}+\dfrac{3\sqrt{x}-2}{x-4}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{2\sqrt{x}-x}\right)\)

ĐKXĐ: x>0; x<>4

\(A=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)+3\left(\sqrt{x}+2\right)+3\sqrt{x}-2}{x-4}:\dfrac{x+3\sqrt{x}-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-2\sqrt{x}+3\sqrt{x}+6+3\sqrt{x}-2}{x-4}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x+\sqrt{x}}\)

\(=\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{x+4\sqrt{x}+4}{x+3\sqrt{x}+2}\)

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)