\(\sqrt{1+n^2+\frac{n^2}{\left(n+1\right)^2}}=\sqrt{\frac{n^2\left(n+1\right)^2+n^2+\left(n+1\right)^2}{\left(n+1\right)^2}}=\sqrt{\frac{n^2\left(n+1\right)^2+2n^2+2n+1}{\left(n+1\right)^2}}\)
\(=\sqrt{\frac{n^2\left(n+1\right)^2+2n\left(n+1\right)+1}{\left(n+1\right)^2}}=\sqrt{\frac{\left[n\left(n+1\right)+1\right]^2}{\left(n+1\right)^2}}=\frac{n\left(n+1\right)+1}{n+1}=n+\frac{1}{n+1}\)
Thay \(n=2014\Rightarrow B=2014+\frac{1}{2015}+\frac{2014}{2015}=2015\)
Câu 2:
\(2\left(x^2+2x+1\right)+3y^2=21\)
\(\Leftrightarrow2\left(x+1\right)^2+3y^2=21\)
Do \(2\left(x+1\right)^2\ge0\Rightarrow3y^2\le21\Rightarrow y^2\le7\)
Mà \(2\left(x+1\right)^2\) chẵn \(\Rightarrow3y^2\) lẻ \(\Rightarrow y^2\) lẻ
\(\Rightarrow y^2=1\Rightarrow y=\pm1\)
\(\Rightarrow2\left(x+1\right)^2+1=21\Rightarrow\left(x+1\right)^2=10\Rightarrow\) không tồn tại x thỏa mãn
Vậy pt trên ko có nghiệm nguyên
