S=\(\frac{1}{5.6}\)+\(\frac{1}{10.9}\)+\(\frac{1}{15.12}\)+....+\(\frac{1}{3350.2013}\) Tính S
Giúp với
tính A= \(\frac{1}{5.6}+\frac{1}{10.9}+\frac{1}{15.12}+.............+\frac{1}{3350.2013}\)
GIÚP MÌNH VỚI NHÉ
Tính S biết
\(S=\frac{15}{12.17}+\frac{35}{17.38}-\frac{39}{18.21}+\frac{30}{21.72}\\ S=\frac{1}{5.6}+\frac{1}{10.9}+\frac{1}{15.12}+...+\frac{1}{3350.2013}\)
Giúp mk vs m.n ơi
mk cần gấp lắm
Thanks m.n nhìu ^^
\(B=\frac{1}{5.6}+\frac{1}{10.9}+\frac{1}{15.12}+...+\frac{1}{3350.2013}\)
\(B=\frac{1}{5.3}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{670.671}\right)\)
\(B=\frac{1}{15}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{670}-\frac{1}{671}\right)\)
\(B=\frac{1}{15}.\left(1-\frac{1}{671}\right)\)
\(B=\frac{1}{15}.\frac{670}{671}=\frac{134}{2013}\)
Nguyễn Huy Thắngsoyeon_Tiểubàng giảiSilver bulletLê Nguyên HạoPhương AnVõ Đông Anh Tuấnsoyeon_Tiểubàng giảiLê Thị Linh ChiNguyễn Huy Tú
Tính tổng S = \(\frac{1}{5.6}+\frac{1}{9.10}+\frac{1}{15.12}+.....+\frac{1}{3350.2013}\)
s=\(\frac{1}{5.3.2}\) +\(\frac{1}{5.3.2.3}\) +.............+\(\frac{1}{5.3.670.671}\)
s=1/15(1/1.2+1/2.3+..................+1/670.671)
s=1/15(1-1/2+1/2-1/3+.............+1/670-1/671)
s=1/15(1-1/671)
s=1/15.670/671
s=134/2013
Tính B= \(\left(\frac{3}{429}-\frac{1}{1.3}\right)\left(\frac{3}{429}-\frac{1}{3.5}\right)....\left(\frac{3}{429}-\frac{1}{119.121}\right)\left(\frac{3}{429}-\frac{1}{121.123}\right)\)
Tính C= \(\frac{1}{5.6}+\frac{1}{10.9}+\frac{1}{15.12}+...+\frac{1}{3350.2013}\)
Tính tổng sau:
S= 1/5.6 + 1/10.9 + 1/15.12+...... + 1/3350.2013
thi thoang giup ti
S = 1/5.6 + 1/10.9+....+ 1/3350.2013
=1/5 . 1/3 .( 1/2+ 1/2.3 + 1/3.4 +... + 1/670.671)
=1/15. ( 1-1/2 + 1/2 - 1/3+...+ 1/670-1/671)
= 1/15 .( 1 - 1/671 )
= 1/15 .670/671
=134/2013
Cám ơn tôi đê. đề cô Phượng đúng không
tính tổng S=1/5.6+1/10.9+1/15.12+-----------+1/3350.2013
đố ai làm được :o
\(S=\dfrac{1}{5.6}+\dfrac{1}{10.9}+\dfrac{1}{15.12}+...+\dfrac{1}{3350.2013}\\ \Rightarrow S=\dfrac{1}{5.3}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{670.671}\right)\\ \Rightarrow S=\dfrac{1}{5.3}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{670}-\dfrac{1}{671}\right)\\ \Rightarrow S=\dfrac{1}{5.3}\left(1-\dfrac{1}{671}\right)\\ \Rightarrow S=\dfrac{1}{5.3}.\dfrac{670}{671}\\ \Rightarrow S=\dfrac{1}{15}.\dfrac{670}{671}=\dfrac{134}{2013}\)
tính tổng S= \(\frac{1}{5.6}\)+\(\frac{1}{10.9}\)+\(\frac{1}{15.12}\)+....+\(\frac{1}{3350.2013}\)
cho biểu thức S= \(\frac{2}{10.12}\)+\(\frac{2}{12.14}\)+\(\frac{2}{14.16}\)+...+\(\frac{2}{98.100}\). Chứng minh S < \(\frac{1}{10}\)
S=1/5.6+1/10.9+1/15.12+...+1/3350.2013
=(1/5).(1/3).(1/1.2+1/2.3+1/3.4+...+1/670.671)
=(1/15). (1-1/2+1/2-1/3+...+1/670-1/671)
=(1/15). (1-1/671)
=1/15.670/671
=134/2013
Tính
a)S1=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
b)S2=\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
c)S3=\(\frac{1}{10.9}+\frac{1}{18.13}+\frac{1}{26.17}+...+\frac{1}{802.405}\)
\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)
\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)
\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)
\(\frac{1}{10.9}+\frac{1}{18.13}+\frac{1}{26.17}+...+\frac{1}{802.405}\)
Tính
Đặt Biểu thức trên là A
\(A=\frac{1}{2.5.9}+\frac{1}{2.9.13}+\frac{1}{2.13.17}+...+\frac{1}{2.397.401}+\frac{1}{2.401.405}\)
\(A=\frac{1}{2}\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{397.401}+\frac{1}{401.405}\right)\)
\(4A=\frac{1}{2}\left(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{397.401}+\frac{4}{401.405}\right)\)
\(4A=\frac{1}{2}\left(\frac{9-5}{5.9}+\frac{13-9}{9.13}+\frac{17-13}{13.17}+...+\frac{401-397}{397.401}+\frac{405-401}{401.405}\right)\)
\(4A=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{405}\right)=\frac{1}{2}.\frac{80}{405}=\frac{40}{405}\Rightarrow A=\frac{40}{4.405}=\frac{2}{81}\)