Ta có:

\(\sqrt{\frac{1}{1^2}+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{\frac{\left(1+n+n^2\right)^2}{n^2\left(n+1\right)^2}}\)

\(=\frac{1+n+n^2}{n\left(n+1\right)}=1+\frac{1}{n}-\frac{1}{n+1}\)

Áp dụng bài toán được

\(A=\sqrt{\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{\frac{1}{1^2}+\frac{1}{2021^2}+\frac{1}{2022^2}}\)

\(=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}+\frac{1}{4}+...+1+\frac{1}{2021}-\frac{1}{2022}\)

\(=2020+\frac{1}{2}-\frac{1}{2022}=\)

\(\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2+b^2}}\right):\frac{b}{a-\sqrt{a^2-b^2}}\left(a>b>0\right)\)

\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{\sqrt{a^2-b^2}+a}{\sqrt{a^2-b^2}}.\frac{a-\sqrt{a^2-b^2}}{b}\)

\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-\left(a^2-b^2\right)}{b\sqrt{a^2-b^2}}\)

\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-a^2+b^2}{b\sqrt{a^2-b^2}}\)

\(=\frac{a}{\sqrt{\left(a-b\right)\left(a+b\right)}}-\frac{b^2}{b\sqrt{\left(a-b\right)\left(a+b\right)}}\)

\(=\frac{a-b}{\sqrt{\left(a-b\right)\left(a+b\right)}}\)

\(=\frac{\sqrt{a-b}}{\sqrt{a+b}}\)

theo định lý pi-ta-go ta có : 

\(BC=\sqrt{12^2+16^2}=20cm\)

áp dụng hệ thức lượng ta có 

\(AB^2=BH.BC\)

=> \(BH=\frac{AB^2}{BC}=\frac{12^2}{20}=\frac{144}{20}\)= 7.2 cm

=>\(CH=BC-BH=20-7,2=12,8cm\)