Question

Rút gọn B
\(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)

Help

Answer 1

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

`B=(\sqrtx-1)/(\sqrtx)+(2\sqrtx+1)/(\sqrtx(\sqrtx+1))`

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

`B=(x+2\sqrtx)/(\sqrtx(\sqrtx+1))`

`B=(\sqrtx(\sqrtx+2))/(\sqrtx(\sqrtx+1))`

`B=(\sqrtx+2)/(\sqrtx+1)`

Answer 2

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

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