Question

cho biểu thức P =\(\left(\dfrac{\sqrt{x}}{\sqrt{x-1}}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\sqrt{x}+x}\right)\)với 0<x≠1.

a) Rút gọn P.

b)Tìm  x để P >2

Answer 1

`P=(sqrtx/(sqrtx-1)+sqrtx/(x-1)):(2/x-(2-x)/(xsqrtx+x))`

`đk:x>0,x ne 1`

`P=((x+sqrtx+sqrtx)/(x-1)):(2/x+(x-2)/(x(sqrtx+1)))`

`=(x+2sqrtx)/(x-1):((2sqrtx+2+x-2)/(x(sqrtx+1)))`

`=(x+2sqrtx)/(x-1):(x+2sqrtx)/(x(sqrtx+1))`

`=(x+2sqrtx)/(x-1)*(x(sqrtx+1))/(x+2sqrtx)`

`=(x(sqrtx+1))/((sqrtx-1)(sqrtx+1))`

`=x/(sqrtx-1)`

`b)P>2`

`<=>x/(sqrtx-1)-2>0`

`<=>(x-2sqrtx+2)/(sqrtx-1)>0`

`<=>((sqrtx-1)^2+1)/(sqrtx-1)>0`

`<=>sqrtx-1>0`

`<=>x>1`

Answer 2

a) đk: x>0;x khác 1;0

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

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

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

= \(\dfrac{x}{\sqrt{x}-1}\)

b) Để P > 2

<=> \(\dfrac{x}{\sqrt{x}-1}-2>0\)

<=> \(\dfrac{x-2\sqrt{x}+2}{\sqrt{x}-1}>0\)

<=> \(\dfrac{\left(\sqrt{x}-1\right)^2+1}{\sqrt{x}-1}>0\)

<=> \(\sqrt{x}-1>0\)

<=> x > 1