\(a,x=49\Rightarrow\sqrt{x}=7\Rightarrow\frac{1}{\sqrt{x}-1}=\frac{1}{6}\)
\(b,x=4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}\right)^2-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\Rightarrow=\sqrt{x}=\sqrt{3}-1\) \(\Rightarrow B=\frac{1}{\sqrt{3}-2}\)
\(c,A=\frac{x+2}{\left(\sqrt{x}\right)^3-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\Rightarrow A-B=\frac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) \(=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}.xét:x+\sqrt{x}+1-3\sqrt{x}=x-2\sqrt{x}+1=\left(\sqrt{x}-1\right)^2\ge0\Rightarrow S\le\frac{1}{3}.\text{Dâu "=" xay}\Leftrightarrow x=1\left(loạidođkxd\right)\Rightarrow S< \frac{1}{3}\)
