a, Ta có : \(A=\left(\frac{x-\sqrt{x}+2}{x-1}-\frac{1}{\sqrt{x}-1}\right).\frac{x+2\sqrt{x}}{2x-2\sqrt{x}}\)
=> \(A=\left(\frac{x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\frac{x+2\sqrt{x}}{2x-2\sqrt{x}}\)
=> \(A=\left(\frac{x-\sqrt{x}+2-\left(\sqrt{x}+1\right)}{x-1}\right).\frac{x+2\sqrt{x}}{2x-2\sqrt{x}}\)
=> \(A=\left(\frac{x-2\sqrt{x}+1}{x-1}\right).\frac{x+2\sqrt{x}}{2x-2\sqrt{x}}\)
=> \(A=\left(\frac{\left(\sqrt{x}-1\right)^2}{x-1}\right).\frac{x+2\sqrt{x}}{2x-2\sqrt{x}}\)
=> \(A=\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\frac{x+2\sqrt{x}}{2x-2\sqrt{x}}\)
=> \(A=\frac{\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)}\frac{\left(x+2\sqrt{x}\right)}{\left(2x-2\sqrt{x}\right)}\)
=> \(A=\frac{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}+1\right)\left(2x-2\sqrt{x}\right)}\)
=> \(A=\frac{\left(\sqrt{x}-1\right)\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)2\sqrt{x}\left(\sqrt{x}-1\right)}\)
=> \(A=\frac{\sqrt{x}+2}{2\sqrt{x}+2}\)
b, Ta có : \(A=\frac{\sqrt{x}+1+1}{2\left(\sqrt{x}+1\right)}=\frac{1}{2}+\frac{1}{2\left(\sqrt{x}+1\right)}\)
- Ta thấy : \(\sqrt{x}+1>0\)
=> \(\frac{1}{2\left(\sqrt{x}+1\right)}>0\)
=> \(\frac{1}{2\left(\sqrt{x}+1\right)}+\frac{1}{2}>\frac{1}{2}\)
=> \(A>\frac{1}{2}\) ( đpcm )
