Ta thấy:
\(\sqrt{\dfrac{1-y}{y}}\times\sqrt{\dfrac{y}{1-y}}=1\left(const\right)\)
=> Ta có thể đặt \(\sqrt{\dfrac{1-y}{y}}=t\left(t\ge0\right)\)
\(\Rightarrow\sqrt{\dfrac{y}{1-y}}=\dfrac{1}{t}\)
~ ~ ~
\(\sqrt{\dfrac{1-y}{y}}=t\)
\(\Rightarrow\dfrac{1-y}{y}=t^2\)
\(\Leftrightarrow1-y=yt^2\)
\(\Leftrightarrow yt^2+y=1\)
\(\Leftrightarrow y\left(t^2+1\right)=1\)
\(\Leftrightarrow y=\dfrac{1}{t^2+1}\)
~ ~ ~
\(x=\dfrac{1}{2}\left(t-\dfrac{1}{t}\right)=\dfrac{t^2-1}{2t}\)
\(\Rightarrow x^2+1=\dfrac{\left(t^2-1\right)^2}{4t^2}+1=\dfrac{\left(t^2-1\right)^2+4t^2}{4t^2}=\dfrac{\left(t^2+1\right)^2}{4t^2}\)
\(\Rightarrow\sqrt{x^2+1}=\left|\dfrac{t^2+1}{2t}\right|=\dfrac{t^2+1}{2t}\left(t\ge0\right)\)
~ ~ ~
\(B=\dfrac{2y\sqrt{1+x^2}}{\sqrt{1+x^2}-x}\)
\(=\dfrac{2\times\dfrac{1}{t^2+1}\times\dfrac{t^2+1}{2t}}{\dfrac{t^2+1}{2t}-\dfrac{t^2-1}{2t}}\)
\(=\dfrac{\dfrac{1}{t}}{\dfrac{2}{2t}}=1\)
