Question

rút gọn B

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

Answer 1

ĐKXĐ: \(x>0\)

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

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

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