a) điều kiện xác định : \(x>0;x\ne1\)
ta có : \(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)^2}\)
\(\Leftrightarrow A=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{1}{\sqrt{x}+1}\)
\(\Leftrightarrow A=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\left(\sqrt{x}+1\right)=\dfrac{\left(\sqrt{x}+1\right)^2}{x-\sqrt{x}}\)
b) thay \(x=6+2\sqrt{5}\) vào \(A\) ta có :
\(A=\dfrac{\left(\sqrt{6+2\sqrt{5}}+1\right)^2}{6+2\sqrt{5}-\sqrt{6+2\sqrt{5}}}=\dfrac{\left(\sqrt{\left(\sqrt{5}+1\right)^2}+1\right)^2}{6+2\sqrt{5}-\sqrt{\left(\sqrt{5}+1\right)^2}}\)
\(=\dfrac{\left(2+\sqrt{5}\right)^2}{5+\sqrt{5}}=\dfrac{9+2\sqrt{5}}{5+\sqrt{5}}\)
tới đây mk nghỉ đề sai rồi bn à
