điều kiện xác định : \(x\ge0;x\ne1\)
a) ta có : \(A=\left(\dfrac{1}{1-\sqrt{x}}+\dfrac{1}{1+\sqrt{x}}\right):\left(\dfrac{1}{1-\sqrt{x}}-\dfrac{1}{1+\sqrt{x}}\right)+\dfrac{1}{1-\sqrt{x}}\)
\(\Leftrightarrow A=\left(\dfrac{2}{1-x}\right):\left(\dfrac{2\sqrt{x}}{1-x}\right)+\dfrac{1}{1-\sqrt{x}}\)
\(\Leftrightarrow A=\left(\dfrac{2}{1-x}\right)\left(\dfrac{1-x}{2\sqrt{x}}\right)+\dfrac{1}{1-\sqrt{x}}=\dfrac{1}{\sqrt{x}}+\dfrac{1}{1-\sqrt{x}}\)ta có : \(x=7+4\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{7+4\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)
\(\Rightarrow A=\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{1-2-\sqrt{3}}=\dfrac{5-3\sqrt{3}}{2}\)
b) áp dụng cauchuy-schwarz dạng engel ta có :
\(A=\dfrac{1}{\sqrt{x}}+\dfrac{1}{1-\sqrt{x}}\ge4\)
dấu "=" xảy ra khi : \(\sqrt{x}=1-\sqrt{x}\Leftrightarrow2\sqrt{x}=1\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\)
vậy ....................................................................................................................