a) A = \(\frac{a+1-2\sqrt{a}}{a+1}:\left(\frac{1}{\sqrt{a}+1}-\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}+1\right)}\right)\)
= \(\frac{\left(\sqrt{a}-1\right)^2}{a+1}:\frac{a+1-2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}+1\right)}\)
= \(\frac{\left(\sqrt{a}-1\right)^2}{a+1}.\frac{\left(a+1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)^2}\) = \(\sqrt{a}+1\)
b) a = \(2020-2.\sqrt{2019}\) = \(\left(\sqrt{2019}-1\right)^2\)
=> \(A=\sqrt{\left(\sqrt{2019}-1\right)^2}+1\) = \(\sqrt{2019}\)
a) Ta có: \(A=\left(1-\frac{2\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}+1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}+a+1}\right)\)
\(=\frac{a+1-2\sqrt{a}}{a+1}:\left(\frac{a+1}{\left(\sqrt{a}+1\right)\left(a+1\right)}-\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\frac{\left(\sqrt{a}-1\right)^2}{a+1}:\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(a+1\right)}\)
\(=\frac{\left(\sqrt{a}-1\right)^2}{a+1}\cdot\frac{\left(\sqrt{a}+1\right)\left(a+1\right)}{\left(\sqrt{a}-1\right)^2}\)
\(=\sqrt{a}+1\)
b) ĐKXĐ: \(a\ge0\)
Ta có: \(a=2020-2\sqrt{2019}\)
\(=2019-2\cdot\sqrt{2019}\cdot1+1\)
\(=\left(\sqrt{2019}-1\right)^2\)
Thay \(a=\left(\sqrt{2019}-1\right)^2\) vào biểu thức \(A=\sqrt{a}+1\), ta được:
\(A=\sqrt{\left(\sqrt{2019}-1\right)^2}+1\)
\(=\sqrt{2019}-1+1\)
\(=\sqrt{2019}\)
Vậy: \(\sqrt{2019}\) là giá trị của biểu thức \(A=\left(1-\frac{2\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}+1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}+a+1}\right)\) tại \(a=2020-2\sqrt{2019}\)
