4.
a) Điều kiện xác định: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}+2\ne0\\\sqrt{x}-2\ne0\\x-4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
\(A=\dfrac{4}{\sqrt{x}+2}+\dfrac{2}{\sqrt{x}-2}-\dfrac{5\sqrt{x}-6}{x-4}\)
\(=\dfrac{4\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{5\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{4\left(\sqrt{x}-2\right)+2\left(\sqrt{x}+2\right)-\left(5\sqrt{x}-6\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{4\sqrt{x}-8+2\sqrt{x}+4-5\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{1}{\sqrt{x}-2}\)
b) Thay \(x=\dfrac{1}{9}\) (thỏa mãn điều kiện xác định) vào biểu thức \(A,\) ta có:
\(A=\dfrac{1}{\sqrt{\dfrac{1}{9}}-2}=\dfrac{1}{\dfrac{1}{3}-2}=\dfrac{1}{\dfrac{1-6}{3}}=\dfrac{1}{\dfrac{-5}{3}}=-\dfrac{3}{5}\)
3.
b) \(\sqrt{7-4\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{4-4\sqrt{3}+3}-\sqrt{3+2\sqrt{3}+1}\)
\(=\sqrt{2^2-2.2\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2+2.1.\sqrt{3}+1^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left|2-\sqrt{3}\right|-\left|\sqrt{3}+1\right|\)
\(=2-\sqrt{3}-\sqrt{3}-1\) (do \(2-\sqrt{3}>0,\) \(\sqrt{3}+1>0\))
\(=1-2\sqrt{3}\)
