a) điều kiện xác định : \(x>0;x\ne1\)
ta có : \(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(1-\dfrac{1}{\sqrt{x}}\right)\)
\(\Leftrightarrow A=\left(\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)
\(\Leftrightarrow A=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)=\dfrac{2}{\sqrt{x}+1}\)
b) ta có : \(x=3-2\sqrt{2}\Rightarrow\sqrt{x}=\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)
\(\Rightarrow A=\dfrac{2}{\sqrt{2}-1+1}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
c) ta có : \(xA=\dfrac{8}{3}\Leftrightarrow\dfrac{2x}{\sqrt{x}+1}=\dfrac{8}{3}\Leftrightarrow6x=8\sqrt{x}+8\)
\(\Leftrightarrow6x-8\sqrt{x}-8=0\Leftrightarrow\left(\sqrt{x}-2\right)\left(6\sqrt{x}+4\right)=0\)
\(\Leftrightarrow\sqrt{x}-2=0\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\) vậy \(x=4\)
