\(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}\left(a>0\right)\)
=\(\sqrt{\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}}=\sqrt{\frac{a^2\left(a^2+2a+1\right)+a^2+2a+1+a^2}{a^2\left(a+1\right)^2}}=\sqrt{\frac{a^4+2a^3+a^2+2a^2+2a+1}{a^2\left(a+1\right)^2}}\)
=\(\sqrt{\frac{a^4+2a^3+3a^2+2a+1}{a^2\left(a+1\right)^2}}=\sqrt{\frac{\left(x^4+2x^2+1\right)+2a\left(a^2+1\right)+a^2}{a^2\left(a+1\right)^2}}=\sqrt{\frac{\left(a^2+1\right)^2+2a\left(a^2+1\right)+a^2}{a^2\left(a+1\right)^2}}\)
=\(\sqrt{\frac{\left(a^2+1+a\right)^2}{a^2\left(a+1\right)^2}}\)=\(\left|\frac{a^2+a+1}{a\left(a+1\right)}\right|=\frac{a^2+a+1}{a\left(a+1\right)}\)(do a>0)
=\(\frac{a\left(a+1\right)+1}{a\left(a+1\right)}=1+\frac{1}{a\left(a+1\right)}=1+\frac{1}{a}-\frac{1}{a+1}\)
=> \(A=1+\frac{1}{a}-\frac{1}{a+1}\)
Áp dụng A vào B có:
B=\(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}\)=\(1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{99}-\frac{1}{100}\)=\(100-\frac{1}{100}\)=\(99,99\)
