\(a,x=4-2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}-1\\ \Leftrightarrow A=\dfrac{\sqrt{3}+1}{\sqrt{3}}=\dfrac{3+\sqrt{3}}{3}\\ b,B=\dfrac{x+12+\sqrt{x}-2-4\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ B=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\\ c,A\cdot B=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\left(\dfrac{2}{\sqrt{x}+1}>0\right)\)
a) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}=\dfrac{\sqrt{4-2\sqrt{3}}+2}{\sqrt{4-2\sqrt{3}}+1}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}+2}{\sqrt{\left(\sqrt{3}-1\right)^2}+1}=\dfrac{\sqrt{3}-1+2}{\sqrt{3}-1+1}=\dfrac{\sqrt{3}+1}{\sqrt{3}}=\dfrac{3+\sqrt{3}}{3}\)
b) \(B=\dfrac{x+12+\sqrt{x}-2-4\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
c) \(AB=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}.\dfrac{\sqrt{x}+2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\)
