\(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{19}=\frac{1}{4}+\left(\frac{1}{5}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+...+\frac{1}{19}\right)\) > \(\frac{1}{4}+\left(\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}\right)+\left(\frac{1}{19}+...+\frac{1}{19}\right)\)> \(\frac{1}{4}+\frac{5}{9}+\frac{10}{19}>\frac{1}{4}+\frac{1}{2}+\frac{1}{2}=1\)

Vậy \(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{19}>1\)

\(B=\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{11}\right)+\left(\frac{1}{12}+...+\frac{1}{19}\right)>\left(\frac{1}{12}+\frac{1}{12}+...+\frac{1}{12}\right)+\left(\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\right)\)=> \(B>\frac{8}{12}+\frac{8}{20}=\frac{2}{3}+\frac{2}{5}=\frac{16}{15}>\frac{15}{15}=1\)

=> ĐPCM

mình có bài làm giống cô Trần Thị Loan

tk mình nhé

\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}>\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)

\(=1-\frac{1}{5}>1\)

                      Kết luận B > 1

Bạn chú ý: Đinh Tuấn Việt đã trả lời sai:

 \(1-\frac{1}{5}\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)(cái này mình cũng ko hiểu sao bạn có thể làm được như vậy)

  nên \(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{19}>\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}

Xét \(\frac{1}{5}\) + \(\frac{1}{6}\) + \(\frac{1}{7}\) +...+\(\frac{1}{12}\) > \(\frac{1}{12}\) x 8= \(\frac{2}{3}\)

Ta có: \(\frac{1}{4}\) = \(\frac{5}{20}\)

Xét \(\frac{1}{13}\) + \(\frac{1}{14}\)+ \(\frac{1}{15}\) +...+ \(\frac{5}{20}\)> \(\frac{1}{20}\) x 8 = \(\frac{2}{5}\)

=> B = \(\frac{1}{4}\) + \(\frac{1}{5}\) + \(\frac{1}{6}\) +...+ \(\frac{1}{19}\) > \(\frac{2}{3}\)+ \(\frac{2}{5}\)= \(\frac{16}{15}\) > 1 => B>1

Ê cậu ghi phân  số kiểu gì thế chỉ cho mình biết cách ấn mình mới giải được

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}\)

    \(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{43}-\frac{1}{46}\)

      \(=1-\frac{1}{46}< 1\)

Vậy \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}< 1\)

1, 3A = 1+1/3 +1/ 3^2 +......+1/3^99                                                                                                                                                        2A = 3A-A =(1+1/3+1/3^2+.....+1/3^99) - (1/3+1/3^2+1/3^3 +.....+1/3^100)                                                                                                   = 1 - 1/3^100                                                                                                                                                                                  A= (1 - 1/3^100) / 2

\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}>\frac{1}{4}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{1}{4}+\frac{15}{20}=1\)

\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+....+\frac{1}{20}+\frac{1}{4}=\frac{3}{4}+\frac{1}{4}=1\)

Vậy B>1

Hok tốt

B=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

help me