Question

Cho biểu thức: A = \(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\) và B = \(\dfrac{1}{\sqrt{x}-2}\) - \(\dfrac{2\sqrt{x}}{x-4}\) + \(\dfrac{4}{\sqrt{x}+2}\) Với x ≥ 0 ; x ≠ 4

Chứng minh B = \(\dfrac{3}{\sqrt{x}+2}\)

Answer 1

\(B=\dfrac{1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{x-4}+\dfrac{4}{\sqrt{x}+2}\)

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

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

\(=\dfrac{-\sqrt{x}+2-4\sqrt{x}+8}{x-4}=\dfrac{-5\sqrt{x}+10}{x-4}\)

\(=\dfrac{-5\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=-\dfrac{5}{\sqrt{x}+2}\)