a: \(P=\left(1-\frac{2\sqrt{a}}{a+1}\right):\left(\frac{1}{1+\sqrt{a}}-\frac{2\sqrt{a}}{a\cdot\sqrt{a}+a+\sqrt{a}+1}\right)\)
\(=\frac{a+1-2\sqrt{a}}{a+1}:\left(\frac{1}{\sqrt{a}+1}-\frac{2\sqrt{a}}{a\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)}\right)\)
\(=\frac{\left(\sqrt{a}-1\right)^2}{a+1}:\left(\frac{1}{\sqrt{a}+1}-\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\frac{\left(\sqrt{a}-1\right)^2}{a+1}:\frac{a+1-2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}+1\right)}=\frac{\left(\sqrt{a}-1\right)^2}{a+1}\cdot\frac{\left(a+1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)^2}\)
\(=\sqrt{a}+1\)
b: \(a=2020-2\cdot\sqrt{2019}\)
\(=2019-2\cdot\sqrt{2019}\cdot1+1=\left(\sqrt{2019}-1\right)^2\)
=>\(P=\sqrt{\left(\sqrt{2019}-1\right)^2}+1=\sqrt{2019}-1+1=\sqrt{2019}\)
