Question

B4: Cho: 

\(M=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{2}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)      với \(x\ge0,x\ne1\)

 a,  Rút gọn M 

 b,  Tính M khi x=9

Answer 1

a. `M=(x+2)/(x\sqrtx-1)+(\sqrt2+1)/(x+\sqrtx+1)-1/(\sqrtx-1)`

`=(x+2)/( (\sqrtx)^3 -1^3))+(\sqrt2+1)/(x+\sqrtx+1)-1/(\sqrtx-1)`

`= (x+2)/((\sqrtx-1)(x+\sqrtx+1)) + +(\sqrt2+1)/(x+\sqrtx+1)-1/(\sqrtx-1)`

`= ((x+2) +(\sqrt2+1)(\sqrtx-1)-(x+\sqrtx+1))/((\sqrtx-1)(x+\sqrtx+1))`

`=( \sqrt2 (\sqrtx-1))/((\sqrtx-1)(x+\sqrtx+1))`

`= (\sqrt2)/(x+\sqrtx+1)`

b. `x=9 => M=\sqrt2/(9+\sqrt9+1)=\sqrt2/13`

Answer 2

a) Ta có: \(M=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)

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

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

b) Thay x=9 vào M, ta được:

\(M=\dfrac{3}{9+3+1}=\dfrac{3}{13}\)