Question

Cho biểu thức:

A = \(\left(\dfrac{1}{x+2\sqrt{x}}-\dfrac{1}{\sqrt{x}+2}\right)\div\dfrac{1-\sqrt{x}}{x+4\sqrt{x}+4}\) với x > 0, x ≠ 1

a) Rút gọn biểu thức A.

b) Tìm x để A = \(\dfrac{5}{3}\)

 

Answer 1

`(a):\  A=((1)/(x+2\sqrt{x})-(1)/(\sqrt{x}+2)):(1-\sqrt{x})/(x+4\sqrt{x}+4)`

`(ĐK:x>0;x\ne 1)`

`=((1)/(\sqrt{x}(\sqrt{x}+2))-(1)/(\sqrt{x}+2)).((\sqrt{x}+2)^{2})/(1-\sqrt{x})`

`=(1-\sqrt{x})/(\sqrt{x}(\sqrt{x}+2)).((\sqrt{x}+2)^{2})/(1-\sqrt{x})`

`=(\sqrt{x}+2)/(\sqrt{x})`

`(b):\ A=5/3<=>(\sqrt{x}+2)/(\sqrt{x})=5/3`

`<=>5\sqrt{x}=3(\sqrt{x}+2)`

`<=>5\sqrt{x}-3\sqrt{x}=6`

`<=>2\sqrt{x}=6`

`<=>\sqrt{x}=3`

`<=>x=9\  (TMDK)`

Answer 2

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

b: Để A=5/3 thì \(\dfrac{\sqrt{x}+2}{\sqrt{x}}=\dfrac{5}{3}\)

=>căn x=3

=>x=9