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 > \(\frac{1}{2}\)
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 > \(\frac{1}{2}\)
chứng minh rằng : \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}< 1\)
CMR\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}<1\)
CMR\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}<1\)
Bài 1: Tính : D =\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
Chứng minh rằng:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}<1\)
CMR
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}<1\)
bài 1:
Cho A= \(\frac{3}{1^2.2^2}\)+ \(\frac{5}{2^2.3^2}\)+ \(\frac{7}{3^2.4^2}\)+.........+ \(\frac{9}{9^2.10^2}\) . So sánh A với 1
Bài 2:
Tìm tổng A= 1 - 7 + 13- 19 + 25 - 31+....... với A có n số hạng
bài 1:
Cho A= \(\frac{3}{1^2.2^2}\)+ \(\frac{5}{2^2.3^2}\)+ \(\frac{7}{3^2.4^2}\)+.........+ \(\frac{9}{9^2.10^2}\) . So sánh A với 1
Bài 2:
Tìm tổng A= 1 - 7 + 13- 19 + 25 - 31+....... với A có n số hạng
Tính
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
Help me...