\(\sqrt{1+\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}}=\sqrt{1+\left(\dfrac{1}{a^2}-\dfrac{2}{a^2\left(a+1\right)^2}+\dfrac{1}{\left(a+1\right)^2}\right)+\dfrac{2}{a^2\left(a+1\right)^2}}\)
\(=\sqrt{1+\dfrac{2}{a^2\left(a+1\right)^2}+\left(\dfrac{1}{a}-\dfrac{1}{a+1}\right)^2}\)
\(=\sqrt{1+\dfrac{2}{\left[a\left(a+1\right)\right]^2}+\dfrac{1}{\left[a\left(a+1\right)\right]^2}}=\sqrt{\left(1+\dfrac{1}{a\left(a+1\right)}\right)^2}\)
\(=\left|1+\dfrac{1}{a^2+1}\right|\)
Do \(a^2\ge0\Leftrightarrow a^2+1>0\Rightarrow\dfrac{1}{a^2+1}>0\Rightarrow\left|1+\dfrac{1}{a^2+1}\right|=1+\dfrac{1}{a^2+1}\)
Nhầm ở kết quả cuối phải là \(\left|1+\dfrac{1}{a\left(a+1\right)}\right|\) nhé
