Tính \(B=\frac{1}{10\cdot9}+\frac{1}{18\cdot13}+\frac{1}{26\cdot17}+.............+\frac{1}{802\cdot405}\)
A=\(\dfrac{1}{10\cdot9}+\dfrac{1}{18\cdot13}+\dfrac{1}{26\cdot17}+...+\dfrac{1}{802\cdot405}\)
nhanh nhé
\(\frac{2}{4\cdot7}-\frac{2}{5\cdot9}+\frac{2}{7\cdot10}-\frac{2}{9\cdot13}+\frac{2}{10\cdot13}-\frac{2}{13\cdot17}+...+\frac{2}{301\cdot304}-\frac{2}{401\cdot405}CM:tich,tren,< \frac{1}{15}\)
\(C=\frac{2}{1\cdot5}+\frac{2}{5\cdot9}+\frac{2}{9\cdot13}+\frac{2}{13\cdot17}+\frac{2}{17\cdot21}\)
\(C=\frac{2}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{17}-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\frac{20}{21}\)
\(=\frac{10}{21}\)
Tinh nhanh:
\(\frac{7}{1\cdot5}+\frac{7}{5\cdot9}+\frac{7}{9\cdot13}+\frac{7}{13\cdot17}+\frac{7}{17\cdot21}\)
= 7/4.(4/1.5 + 4/5.9 + 4/9.13 + 4/13.17 + 4/17.21)
= 7/4.(1-1/5+1/5-1/9+1/9-1/13+1/13-1/17+1/17-1/21)
= 7/4.(1-1/21)
= 7/4.20/21 = 5/3
Tk mk nha
Đặt biểu thức bằng A
4/7A=1-1/5+1/5-1/9+...+1/17_1/21
4/7A=1-1/21
4/7A=20/21
A=35/21=5/3
\(\frac{4}{1\cdot5}\)+\(\frac{4}{5\cdot9}\)+\(\frac{4}{9\cdot13}\)+\(\frac{4}{13\cdot17}\)+\(\frac{4}{17\cdot21}\)< 1
Áp dụng theo dạng toán số ai cập ta có:
4/1.5+4/5.9+4/9.13+4/13.17+4/17.21=1/1-1/5+1/5-1/9+1/9-1/13+1/13-1/17+1/17-1/21=1-1/21 < 1
Vậy tổng đó < 1
\(\text{So sánh: A=\frac{1}{2\cdot3}+\frac{1}{4\cdot5}+\frac{1}{6\cdot7}+\frac{1}{8\cdot9}+\frac{1}{10\cdot11}+\frac{1}{12\cdot13}+\frac{1}{14\cdot15}+\frac{1}{16\cdot17}+\frac{1}{18\cdot19} và B=\frac{9}{19}}\)So sánh: A=1/2*3 + 1/4*5 + 1/6*7 + 1/8*9 + 1/10*11 + 1/12*13 + 1/14*15 + 1/16*17 + 1/18*19 và B=9/19
Giúp tớ với, tớ cần gấp !! Cảm ơn nhìu ạ !!
Ta có
\(C=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}...+\frac{1}{17.18}>A=\frac{1}{2.3}+\frac{1}{5.4}+...+\frac{1}{18.19}\)
\(C< =>\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{18-17}{17.18}\)\(>A\)
\(C< =>\frac{1}{2}-\frac{1}{18}\)\(>A\)
\(C< =>\frac{4}{9}\)\(>A\left(1\right)\)
Lại có \(C=\frac{4}{9}< \frac{9}{19}=B\left(2\right)\)
Từ (1),(2) => B>A
\(\frac{1}{1\cdot3\cdot7}\)+\(\frac{1}{3\cdot7\cdot9}\)+\(\frac{1}{7\cdot9\cdot13}\)+\(\frac{1}{9\cdot13\cdot15}\)+\(\frac{1}{13\cdot15\cdot19}\)
dễ
ta tách ra xog dùng phương pháp loại trừ đó
\(\frac{2}{4\cdot7}\)-\(\frac{3}{5\cdot9}\)+ \(\frac{2}{7\cdot10}\)-\(\frac{3}{9\cdot13}\)+ ....+\(\frac{2}{301\cdot304}\)-\(\frac{3}{401\cdot405}\)
cho \(A=\frac{1}{3\cdot8}+\frac{1}{8\cdot13}+...+\frac{1}{33.38}\)
\(B=\frac{1}{3\cdot10}+\frac{1}{10\cdot17}+...+\frac{1}{31\cdot38}\)
tính tỉ số \(\frac{A}{B}\) \(\left(A⋮B\right)\)
A=\((\frac{1}{3.8}+\frac{1}{8.13}+...+\frac{1}{33.38})\)
A=\(\frac{1}{5}\left(\frac{5}{3.8}+\frac{5}{8.13}+...+\frac{5}{33.38}\right)\)
A=\(\frac{1}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{33}-\frac{1}{38}\right)\)
A=\(\frac{1}{5}.\left(\frac{1}{3}-\frac{1}{38}\right)\)
A=\(\frac{1}{5}.\frac{35}{114}\)
A=\(\frac{7}{114}\)
B=\((\frac{1}{3.10}+\frac{1}{10.17}+...+\frac{1}{31.38})\)
B=\(\frac{1}{7}\left(\frac{7}{3.10}+\frac{7}{10.17}+...+\frac{7}{31.38}\right)\)
B=\(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+...+\frac{1}{31}-\frac{1}{38}\right)\)
B=\(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{38}\right)\)
B=\(\frac{1}{7}.\frac{35}{114}\)
B=\(\frac{5}{114}\)
⇒ \(\frac{A}{B}\)=\(\frac{7}{114}:\frac{5}{114}=\frac{7}{114}.\frac{114}{5}=\frac{7}{5}\)
Vậy \(\frac{A}{B}=\frac{7}{5}\)
A = \(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+....+\frac{1}{33}-\frac{1}{38}\)
=\(\frac{1}{3}-\frac{1}{38}\)
=\(\frac{35}{114}\)
B =\(\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+...+\frac{1}{31}-\frac{1}{38}\)
=\(\frac{1}{3}-\frac{1}{38}\)
=\(\frac{35}{114}\)
=>tỉ số \(\frac{A}{B}\)= \(\frac{35}{114}:\frac{35}{114}\)=1
A=\(\frac{1}{3.8}+\frac{1}{8.13}+...+\frac{1}{33.38}\)
=\(\frac{1}{5}\left(\frac{5}{3.8}+\frac{5}{8.13}+...+\frac{5}{33.38}\right)\)
=\(\frac{1}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{33}-\frac{1}{38}\right)\)
=\(\frac{1}{5}\left(\frac{1}{3}-\frac{1}{38}\right)\)
=\(\frac{1}{5}.\frac{35}{114}\)
=\(\frac{7}{114}\)
B=\(\frac{1}{3.10}+\frac{1}{10.17}+...+\frac{1}{31.38}\)
=\(\frac{1}{7}\left(\frac{7}{3.10}+\frac{7}{10.17}+...+\frac{7}{31.38}\right)\)
=\(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+...+\frac{1}{31}-\frac{1}{38}\right)\)
=\(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{38}\right)\)
=\(\frac{1}{7}.\frac{35}{114}\)
=\(\frac{5}{114}\)
Vì \(\frac{7}{114}>\frac{5}{114}\)
⇒ A > B
Vậy A > B