Question

[shunnokeshi]

rút gọn biểu thức A=\(\frac{1}{\sqrt{3+2\sqrt{2}}}\)+\(\frac{1}{\sqrt{7+2\sqrt{12}}}\)+...+\(\frac{1}{\sqrt{2x+1+2\sqrt{n^2+n}}}\)trong đó n=2018

Answer 1

[Đoàn Đức Hà]

\(\frac{1}{\sqrt{2k+1+2\sqrt{k^2+k}}}=\frac{1}{\sqrt{k+1+2\sqrt{k\left(k+1\right)}+k}}=\frac{1}{\sqrt{k+1}+\sqrt{k}}\)

\(=\frac{\sqrt{k+1}-\sqrt{k}}{\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{k+1-k}=\sqrt{k+1}-\sqrt{k}\)

Do đó ta có: 

\(A=\frac{1}{\sqrt{3+2\sqrt{2}}}+...+\frac{1}{\sqrt{2n+1+2\sqrt{n^2+n}}}\)

\(A=\sqrt{2}-\sqrt{1}+...+\sqrt{n+1}-\sqrt{n}\)

\(A=\sqrt{n+1}-1\)

Với \(n=2018\)ta có: \(A=\sqrt{2019}-1\).