Chứng tỏ:
\(\frac{1}{6}+\frac{1}{7}+....+\frac{1}{19}< 2\)
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\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}>\frac{1}{4}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{1}{4}+\frac{15}{20}=1\)
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+....+\frac{1}{20}+\frac{1}{4}=\frac{3}{4}+\frac{1}{4}=1\)
Vậy B>1
Hok tốt
B = \(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\)
B = \(\left(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{11}\right)+\left(\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}\right)>\left(\frac{1}{11}+...+\frac{1}{11}\right)+\left(\frac{1}{19}+...+\frac{1}{19}\right)\)
B > \(\frac{240}{209}\)
Vậy B > 1.
B=1/4+(1/5+1/6+...+1/19)>1/4+15x1/20
B>1/4+15/20=1/4+3/4=1
\(\Rightarrow\)B>1
Đặt A=\(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{17}\)\(=\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{10}\right)+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{17}\right)\)
Có: \(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{10}< \frac{1}{5}.6=\frac{6}{5}\)(1)
\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{17}< \frac{1}{11}.7=\frac{7}{11}\)(2)
Từ (1) và (2) suy ra: A\(< \frac{6}{5}+\frac{7}{11}=\frac{101}{55}\)
Lại có: \(\frac{101}{55}< \frac{110}{55}=2\)
Suy ra: A<2 (đpcm)
A=3 /1^2.2^2 +5 / 2^2.3^2 +7/3^2.4^2 +...+ 19 /9^2.10^2
=1/1^2-1/2^2+1/2^2-1/3^2+1/3^2-1/4^2+....+1/9^2-1/10^2
=1/1^2-1/10^2
=99/100
=0,99
vậy A< 1
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}< 1\)
Vậy \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}< 1\)
\(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{19}=\frac{1}{4}+\left(\frac{1}{5}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+...+\frac{1}{19}\right)\) > \(\frac{1}{4}+\left(\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}\right)+\left(\frac{1}{19}+...+\frac{1}{19}\right)\)> \(\frac{1}{4}+\frac{5}{9}+\frac{10}{19}>\frac{1}{4}+\frac{1}{2}+\frac{1}{2}=1\)
Vậy \(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{19}>1\)
Đặt C\(=\frac{1}{6}+\frac{1}{7}+...+\frac{1}{19}\)
\(\)C có 13 phân số tất cả, ta chia ra như sau:
C =1/5+(1/6+....1/11)+(1/12+1/12+.....1/16 +1/17)
Vì trong nhóm I thì 1/ 6 là lớn nhất, nhóm II thì 1/12 là lớn nhất ,xuy ra:
C< 1/5 +6.1/6+6.1/12
C<1/5+ 1 +1/2
C<1+7/10<1+1=2
Vậy C<2
1/6+1/7+...1/19
=(1/6+1/7+...+1/13)+(1/14+1/15+...+1/19)< 7.1/6+6.1/14
=7/6+6/14
=67/42<84/42=2
=> 1/6+1/7+...+1/19<2
k minh nha