Question

Rút gọn biểu thức: \(\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}\)với a>0

Answer 1

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

Answer 2

=\(1+\frac{1}{a}+\frac{1}{a+1}\)

=\(\frac{a+1}{1}-\frac{1}{a+1}\)