Giải câu 5 trang 29 sách phát triển năng lực toán 9 tập 1.
Ta có:
$\left ( 1+\frac{1}{n}-\frac{1}{n+1} \right )^{2}$ = $\left ( \frac{n+1}{n}-\frac{1}{n+1} \right )^{2}$ = $\left ( \frac{n+1}{n}\right )^{2}-2.\frac{n+1}{n}.\frac{1}{n+1}+\left ( \frac{1}{n+1} \right )^{2}$
= $\frac{n^{2}+2n+1}{n^{2}}-2.\frac{1}{n}+\frac{1}{(n+1)^{2}}$ = $1+\frac{2}{n}+\frac{1}{n^{2}}-.\frac{2}{n}+\frac{1}{(n+1)^{2}}=1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}$
=> $\sqrt{1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}}=\sqrt{\left ( 1+\frac{1}{n}-\frac{1}{n+1} \right )^{2}}=\left | 1+\frac{1}{n}-\frac{1}{n+1} \right |$
=> ĐPCM
Từ đó ta có:
S = $\sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+\sqrt{1+\frac{1}{3^{2}}+\frac{1}{4^{2}}}+\sqrt{1+\frac{1}{4^{2}}+\frac{1}{5^{2}}}+...+\sqrt{1+\frac{1}{2018^{2}}+\frac{1}{2019^{2}}}$.
= $\left | 1+\frac{1}{2}-\frac{1}{2+1} \right |+\left | 1+\frac{1}{3}-\frac{1}{3+1} \right |+\left | 1+\frac{1}{4}-\frac{1}{4+1} \right |+...+\left | 1+\frac{1}{2018}-\frac{1}{2018+1} \right |$
= $1.2017+\frac{1}{2}-\frac{1}{2019}$ = $2017+\frac{2017}{2019}$ = $2017.(1+\frac{1}{2019})$ = $\frac{2017.2020}{2019}$