Question

[Nguyễn Thị Khánh Vy]

So sánh

F = \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+......+\frac{19}{9^2.10^2}\)với 1

E = \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}\)với \(\frac{1}{2}\)

Answer 1

[Trần Ngọc Linh]

ta có:

 \(\frac{1}{11}\)>\(\frac{10}{20}\)

\(\frac{1}{12}\)>\(\frac{10}{20}\)

\(\frac{1}{13}\)>\(\frac{10}{20}\)

....

\(\frac{1}{19}\)>\(\frac{10}{20}\)

=>E >\(\frac{10}{20}\)

vậy E > \(\frac{1}{2}\)