Bài 6:
a: \(A=\left(1-\frac{2\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}+1}-\frac{2\sqrt{a}}{a\cdot\sqrt{a}+a+\sqrt{a}+1}\right)\)
\(=\frac{a+1-2\sqrt{a}}{a+1}:\left(\frac{1}{\sqrt{a}+1}-\frac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a+1\right)}\right)\)
\(=\frac{\left(\sqrt{a}-1\right)^2}{a+1}:\frac{a+1-2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a+1\right)}=\frac{\left(\sqrt{a}-1\right)^2}{a+1}\cdot\frac{\left(\sqrt{a}+1\right)\left(a+1\right)}{\left(\sqrt{a}-1\right)^2}=\sqrt{a}+1\)
b: Thay \(a=2021-2\sqrt{2020}=\left(\sqrt{2020}-1\right)^2\) vào A, ta được:
\(A=\sqrt{\left(\sqrt{2020}-1\right)^2}+1=\sqrt{2020}-1+1=\sqrt{2020}\)
Bài 3:
a: \(P=\left(\frac{1}{\sqrt{a}-3}+\frac{1}{\sqrt{a}+3}\right)\left(1-\frac{3}{\sqrt{a}}\right)\)
\(=\frac{\sqrt{a}+3+\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\cdot\frac{\sqrt{a}-3}{\sqrt{a}}=\frac{2\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+3\right)}=\frac{2}{\sqrt{a}+3}\)
b: \(P>\frac12\)
=>\(\frac{2}{\sqrt{a}+3}-\frac12>0\)
=>\(\frac{4-\sqrt{a}-3}{2\left(\sqrt{a}+3\right)}>0\)
=>\(1-\sqrt{a}>0\)
=>\(\sqrt{a}<1\)
=>0<a<1
