\(=\left[\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)^2}\right]:\left(\frac{x+1}{x-1}-\frac{x-1}{x-1}\right)\)

\(=\left[\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)^2}\right]:\frac{2}{x-1}\)

\(=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{x-1}{2}\)

\(=\left[\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{x-1}{2}\)

\(=\frac{x-\sqrt{x}-2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2}\)

\(=\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2}\)

\(=\frac{1}{\sqrt{x}+1}\le1\)

Dấu "=" xảy ra khi x=0

thanks bạn 

b, \(\frac{\sqrt{3}}{2+\sqrt{3}}-\frac{\sqrt{3}}{2-\sqrt{3}}\) = \(\frac{\sqrt{3}\left(2-\sqrt{3}\right)-\sqrt{3}\left(2+\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)=\(\frac{2\sqrt{3}-3-2\sqrt{3}-3}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)=\(\frac{-6}{4-3}\)=-6

c,\(\frac{2}{\sqrt{5}-2}-\frac{2}{\sqrt{5}+2}\)=\(\frac{2\left(\sqrt{5}+2\right)-2\left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)=\(\frac{2\sqrt{5}+4-2\sqrt{5}+4}{\sqrt{5}^2-2^2}\)=\(\frac{8}{1}\)=8

AC= 2.5    BC= 5.5