Question

rút gọn

bài 6

\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac{2}{\sqrt{x}-2}\right):\dfrac{x+4}{\sqrt{x}+2}\) vs \(x\ge0,x\ne4\)

Answer 1

\(P=\dfrac{x-2\sqrt{x}+2\sqrt{x}+4}{x-4}\cdot\dfrac{\sqrt{x}+2}{x+4}\)

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

Answer 2

P= \(\left(\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)+2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right).\dfrac{\sqrt{x}+2}{x+4}\)
  =  \(\left(\dfrac{x-2\sqrt{x}+2\sqrt{x}+4}{x-4}\right).\dfrac{\sqrt{x}+2}{x+4}\)
  =   \(\dfrac{\left(x+4\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)\left(x+4\right)}\)
  =\(\dfrac{1}{\sqrt{x}-2}\)  (với x > 0 và x ≠ 4)

Answer 3

\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac{2}{\sqrt{x}-2}\right):\dfrac{x+4}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)+2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{x+4}{\sqrt{x}+2}\\ =\dfrac{x-2\sqrt{x}+2\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{x+4}{\sqrt{x}+2}\\ =\dfrac{x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}+2}{x+4}\\ =\dfrac{1}{\sqrt{x}-2}\) ĐK : \(x\ge0;x\ne4\)

Answer 4

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