Question

Chứng minh rằng: \(\frac{1}{\sqrt{1}}\)+\(\frac{1}{\sqrt{2}}\)+\(\frac{1}{\sqrt{3}}\)+......+\(\frac{1}{\sqrt{121}}\)>11

Answer 1

\(\frac{1}{\sqrt{1}}< \frac{1}{\sqrt{121}}\)

\(\frac{1}{\sqrt{2}}< \frac{1}{\sqrt{121}}\)

................

\(\frac{1}{\sqrt{121}}=\frac{1}{\sqrt{121}}\)

Suy ra \(\frac{1}{\sqrt{1}}\)+\(\frac{1}{\sqrt{2}}\)+.............+\(\frac{1}{\sqrt{121}}\)<\(\frac{1}{\sqrt{121}}+\frac{1}{\sqrt{121}}+\frac{1}{\sqrt{121}}+......\frac{1}{\sqrt{121}}\)=\(\frac{121}{11}\)=11(đpcm)(vì có 121 chữ số)\(\frac{1}{\sqrt{121}}\))

Answer 2

Khuyển Dạ Xoa : \(\sqrt{1}< \sqrt{121}\Rightarrow\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{121}}\)  chứ?

Answer 3

tth đúng rồi đó,Khuyển Dạ Xoa  ngược dấu rồi! :)

Answer 4

vcl, não hơi có vấn đề :)) thanks tth

\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{121}}>\frac{1}{\sqrt{121}}+\frac{1}{\sqrt{121}}+...+\frac{1}{\sqrt{121}}\left(\text{121 số hạng}\right)\)

\(VT>\frac{121}{\sqrt{121}}=\frac{121}{11}=11\)