#)Giải :
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}>\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)
\(B=1-\frac{1}{5}< 1\)
\(\Leftrightarrow\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}>\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)
\(\Leftrightarrow\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}>\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}< 1\)
\(\Leftrightarrow B< 1\)
#~Will~be~Pens~#
Ta có : \(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\)
\(B=\frac{1}{4}+\left[\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right]+\left[\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}\right]\)
Vì \(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}=\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}=\frac{5}{9}>\frac{1}{2}\)
\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}=\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}=\frac{10}{19}>\frac{1}{2}\)
\(\Rightarrow B>\frac{1}{4}+\frac{5}{9}+\frac{10}{19}\)
\(\Rightarrow B>\frac{1}{4}+\frac{1}{2}+\frac{1}{2}\)
\(\Rightarrow B>\frac{1}{4}+\frac{2}{4}+\frac{2}{4}\)
\(\Rightarrow B>\frac{5}{4}>\frac{4}{4}=1\)
Vậy B > 1
bn pe gì đó, B > 1 mà
mk nghĩ là
tất cả các phân số đều lớn hơn 1 = b
nên B lớn hơn 1
#)Úi chết cho mk xin lỗi mk gõ lộn rùi :v , ph là \(B>1\)nha, mong mọi ng bỏ qua
\(\text{Bài giải}\)
\(\text{Ta có : }\)
\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}>\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}\)
\(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}=\frac{1}{3}\)
\(\frac{1}{9} +\frac{1}{10} +\frac{1}{11}>\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)
\(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}>\frac{1}{15}+\frac{1}{15}+\frac{1}{15}=\frac{3}{15}=\frac{1}{5}\)
\(\frac{1}{15}+\frac{1}{16}+\frac{1}{17}>\frac{1}{18}+\frac{1}{18}+\frac{1}{18}=\frac{3}{18}=\frac{1}{6}\)
\(\Rightarrow\text{ }B>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{29}{20}>1\)
\(\Rightarrow\text{ }B>1\)
\(\text{Bài giải}\)
\(\text{Ta có : }\)
\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}>\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}\)
\(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}=\frac{1}{3}\)
\(\frac{1}{9} +\frac{1}{10} +\frac{1}{11}>\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)
\(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}>\frac{1}{15}+\frac{1}{15}+\frac{1}{15}=\frac{3}{15}=\frac{1}{5}\)
\(\frac{1}{15}+\frac{1}{16}+\frac{1}{17}>\frac{1}{18}+\frac{1}{18}+\frac{1}{18}=\frac{3}{18}=\frac{1}{6}\)
\(\Rightarrow\text{ }B>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{29}{20}>1\)
\(\Rightarrow\text{ }B>1\)