ĐK: \(x>0\)
\(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{x+\sqrt{x}}=\frac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\)\(\frac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{x+\sqrt{x}}\)(ĐKXĐ: \(x\ne0\))
\(=\frac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
=\(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
=\(\frac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
=\(\frac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)=\(\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)=\(\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
Vay ket qua rut gon la \(\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
