A=\(\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}\left(a>0\right)\)
=\(\sqrt{\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}}\)
= \(\sqrt{\frac{a^2\left(a^2+2a+1\right)+a^2+2a+1+a^2}{a^2\left(a+1\right)^2}}\)=\(\sqrt{\frac{a^4+2a^3+a^2+2a^2+2a+1}{a^2\left(a+1\right)^2}}\)=\(\sqrt{\frac{a^4+2a^3+3a^2+2a+1}{a^2\left(a+1\right)^2}}\)
=\(\sqrt{\frac{\left(a^4+2a^2+1\right)+2a\left(a^2+1\right)+a^2}{a^2\left(a+1\right)^2}}=\sqrt{\frac{\left(a^2+1\right)^2+2a\left(a^2+1\right)+a^2}{a^2\left(a+1\right)^2}}=\sqrt{\frac{\left(a^2+1+a\right)^2}{a^2\left(a+1\right)^2}}\)
=\(\left|\frac{a^2+a+1}{a\left(a+1\right)}\right|=\frac{a^2+a+1}{a\left(a+1\right)}\)(do a>0)=\(\frac{a\left(a+1\right)+1}{a\left(a+1\right)}=1+\frac{1}{a\left(a+1\right)}=1-\frac{1}{a}-\frac{1}{a+1}\)
Vậy A= \(1-\frac{1}{a}-\frac{1}{a+1}\)
làm lại cái đoạn \(1+\frac{1}{a\left(a+1\right)}=1+\frac{1}{a}-\frac{1}{a+1}\) nha
