1.
\(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)
\(\Rightarrow A=\dfrac{\sqrt{\left(2+\sqrt{3}\right)^2}+1}{\sqrt{\left(2+\sqrt{3}\right)^2}-2}=\dfrac{2+\sqrt{3}+1}{2+\sqrt{3}-2}=\dfrac{3+\sqrt{3}}{\sqrt{3}}=\sqrt{3}+1\)
2.
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-2\sqrt{x}-\left(x-1\right)-\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-3}{\sqrt{x}-2}=\dfrac{3}{2-\sqrt{x}}\)
3.
\(\dfrac{B}{A}< -1\Rightarrow\dfrac{3}{2-\sqrt{x}}.\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< -1\)
\(\Rightarrow\dfrac{-3}{\sqrt{x}+1}< -1\Leftrightarrow-3< -\sqrt{x}-1\)
\(\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)
Kết hợp ĐKXĐ \(\Rightarrow0\le x< 4\)
