1) \(P=\left(\frac{\sqrt{b}+1}{\sqrt{b}-1}-\frac{\sqrt{b}-1}{\sqrt{b}+1}+4\sqrt{b}\right).\frac{1}{2b\sqrt{b}}\)
\(=\left(\frac{\left(\sqrt{b}+1\right)^2}{\left(\sqrt{b}+1\right)\left(\sqrt{b}-1\right)}-\frac{\left(\sqrt{b}-1\right)^2}{\left(\sqrt{b}+1\right)\left(\sqrt{b}-1\right)}+\frac{4\sqrt{b}\left(\sqrt{b}+1\right)\left(\sqrt{b}-1\right)}{\left(\sqrt{b}+1\right)\left(\sqrt{b}-1\right)}\right).\frac{1}{2b\sqrt{b}}\)
\(=\frac{\left(b+2\sqrt{b}+1\right)-\left(b-2\sqrt{b}+1\right)+4\sqrt{b}\left(b-1\right)}{\left(\sqrt{b}+1\right)\left(\sqrt{b}-1\right)}.\frac{1}{2b\sqrt{b}}\)
\(=\frac{4\sqrt{b}+4b\sqrt{b}-4\sqrt{b}}{\left(b-1\right)2b\sqrt{b}}=\frac{2}{b-1}\)
2) Để P = b thì :
\(\frac{2}{b-1}=b\Leftrightarrow b\left(b-1\right)=2\)
\(\Leftrightarrow b^2-b-2=0\Leftrightarrow\left(b+1\right)\left(b-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b+1=0\\b-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}b=-1\\b=2\end{matrix}\right.\)
Vậy ...
