a) Để B có nghĩa thì \(\left\{{}\begin{matrix}y\ge0\\y\ne1\end{matrix}\right.\)
B=\(\left(\dfrac{1}{\sqrt{y}+1}-\dfrac{3\sqrt{y}}{\sqrt{y}-1}+3\right).\dfrac{\sqrt{y}+1}{\sqrt{y}+2}=\left[\dfrac{\sqrt{y}-1}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}-\dfrac{3\sqrt{y}\left(\sqrt{y}+1\right)}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}+\dfrac{3\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}\right].\dfrac{\sqrt{y}+1}{\sqrt{y}+2}=\left[\dfrac{\sqrt{y}-1}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}-\dfrac{3y+3\sqrt{y}}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}+\dfrac{3y-3}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}\right].\dfrac{\sqrt{y}+1}{\sqrt{y}+2}=\dfrac{\sqrt{y}-1-3y-3\sqrt{y}+3y-3}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}.\dfrac{\sqrt{y}+1}{\sqrt{y}+2}=\dfrac{\left(-2\sqrt{y}-4\right)\left(\sqrt{y}+1\right)}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)\left(\sqrt{y}+2\right)}=\dfrac{-2\left(\sqrt{y}+2\right)\left(\sqrt{y}+1\right)}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)\left(\sqrt{y}+2\right)}=\dfrac{-2}{\sqrt{y}-1}=\dfrac{2}{1-\sqrt{y}}\)
b) Ta có y=\(3+2\sqrt{2}\Rightarrow P=\dfrac{2}{1-\sqrt{3+2\sqrt{2}}}=\dfrac{2}{1-\sqrt{2+2\sqrt{2}+1}}=\dfrac{2}{1-\sqrt{\left(\sqrt{2}+1\right)^2}}=\dfrac{2}{1-\sqrt{2}-1}=\dfrac{2}{-\sqrt{2}}=-\sqrt{2}\)
Vậy khi x=\(3+2\sqrt{2}\) thì \(P=-\sqrt{2}\)
