Question

cho A = \(\left(\dfrac{\sqrt{x+1}}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\right):\dfrac{3\sqrt{x}-x}{x+4\sqrt{x}+4}\)

rut gon A

Answer 1

\(A=\dfrac{-\left(\sqrt{x}+1\right)\left(2+\sqrt{x}\right)-2\sqrt{x}\left(2-\sqrt{x}\right)+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+2\right)^2}\)

\(A=\dfrac{-3\sqrt{x}-x-2-4\sqrt{x}+2x+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)

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

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

 

Answer 2

CHÚC BẠN HỌC TỐT NHÉ

Answer 3

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