Question

[Nguyễn Thị Hà Linh]

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

a) Rút gọn

b) Tính P khi \(x=2019-2\sqrt{2018}\)

Answer 1

[Nhiên An Trần]

a, \(P=\left(1-\dfrac{2\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}+x+1}\right)\)(ĐK: \(x\ge0,x\ne-1\))

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

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

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

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

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

b, ĐK: \(x\ge0,x\ne-1\)

\(x=2019-2\sqrt{2018}=\left(\sqrt{2018}-1\right)^2\)

Thay \(x=\left(\sqrt{2018}-1\right)^2\)(TMĐK) vào P ta có:

\(P=\sqrt{2018}-1+1=\sqrt{2018}\)

Vậy với \(x=2019-2\sqrt{2018}\) thì \(P=\sqrt{2018}\)