a) điều kiện xác định : \(x>0;x\ne1\)
ta có : \(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\left(1+\dfrac{1}{\sqrt{x}}\right)\)
\(\Leftrightarrow B=\left(\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)
\(\Leftrightarrow B=\left(\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)=\dfrac{4}{\sqrt{x}-1}\)
b) ta có : \(x=4+2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(\Rightarrow B=\dfrac{4}{\sqrt{3}+1-1}=\dfrac{4}{\sqrt{3}}\)
c) ta có : \(B=1\Leftrightarrow\dfrac{4}{\sqrt{x}-1}=1\Leftrightarrow\sqrt{x}-1=4\Leftrightarrow\sqrt{x}=5\)
\(\Leftrightarrow x=25\left(tmđk\right)\) vậy \(x=25\)
