A><=1

k mình nha

A= 1/10+1/11+1/12+1/13+...........+1/99+1/100

2A=1/9+1/10+1/11+1/12+...........+1/98+1/99

2A-A=(1/10+1/11+1/12+1/13+.............+1/99+1/100)-(1/9+1/10+1/11+1/12+............1/98+1/99)

A=1/100-1/9

=>A<1

A<1,nhìn là bít liền à !^~^

kb đc 0

2 câu đầu tôi làm đc

a) Ta có :

\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}\)

\(>\frac{1}{10}+\frac{1}{100}.90=\frac{1}{10}+\frac{90}{100}=1\)

vậy A > 1

b) \(B=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\)

\(>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{1}{20}.10=\frac{1}{2}\)

Vậy B > \(\frac{1}{2}\)

lồn

Ta có:

\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)

\(=\frac{1}{4}+\left(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\right)\)

Đặt \(B=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)

\(B=\left(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}\right)+\left(\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\right)\)

Giả sử tất cả các số hạng của B đều bằng \(\frac{1}{6^2}\)

\(\Rightarrow B=6.\frac{1}{6^2}=\frac{6}{36}=\frac{1}{6}<\frac{1}{4}\)

Do đó \(B<\frac{1}{4}\)

\(\Rightarrow A=\frac{1}{4}+B<\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)

Vậy \(A<\frac{1}{2}\)

\(=-\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)

\(=-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\)

\(=-\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{99.101}{100^2}\)

\(=-\frac{1.2....99}{2.3...100}.\frac{3.4....101}{2.3...100}\)

\(=-\frac{1}{100}.\frac{101}{2}=\frac{-101}{200}\)

Học good

\(=-\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)

\(=-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\)

\(=-\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}...\frac{99.101}{100^2}\)

\(=-\frac{1.2...99}{2.3...100}\cdot\frac{3.4...101}{2.3.100}\)

\(=-\frac{1}{100}\cdot\frac{101}{2}\)

\(=-\frac{101}{200}\)

A < 1. mình học rồi, k mình nhé

\(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+....+\frac{1}{20}\)

\(=\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}\right)\)

\(>\frac{1}{15}\cdot5+\frac{1}{20}\cdot5\)

\(=\frac{1}{3}+\frac{1}{4}\)

\(=\frac{7}{12}>\frac{6}{12}=\frac{1}{2}\)

\(\Rightarrow S>\frac{1}{2}\)

Bài làm

Ta có: 

\(\frac{1}{11}>\frac{1}{20}\), \(\frac{1}{12}>\frac{1}{20}\), \(\frac{1}{13}>\frac{1}{20}\), \(\frac{1}{14}>\frac{1}{20}\), \(\frac{1}{15}>\frac{1}{20}\), \(\frac{1}{16}>\frac{1}{20}\), \(\frac{1}{17}>\frac{1}{20}\), \(\frac{1}{18}>\frac{1}{20}\),\(\frac{1}{19}>\frac{1}{20}\)

=> \(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}\)

hay \(\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}\)

=> \(S=\frac{1}{20}.10=\frac{10}{20}=\frac{1}{2}\)

Do đó: \(S=\frac{1}{2}\)

# Chúc bạn học tốt #

Ta có các phân số : \(\frac{1}{11};\frac{1}{12};\frac{1}{13};\frac{1}{14};\frac{1}{15};\frac{1}{16};\frac{1}{17};\frac{1}{18};\frac{1}{19}>\frac{1}{20}\)

Do đó : \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)có 10 phân số \(\frac{1}{20}\)

\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{10}{20}\)

\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{1}{2}\)

Vậy : \(S>\frac{1}{2}\)