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A=1/2.9+1/9.7+1/7.19+...+1/252.509
=?
??????
b)\(\dfrac{1}{7}B=\dfrac{1}{10.18}+\dfrac{1}{18.26}+\dfrac{1}{26.34}+...+\dfrac{1}{802.810}\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{8}{10.18}+\dfrac{8}{18.26}+\dfrac{8}{26.34}+...+\dfrac{8}{802.810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{1}{10}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{34}+...+\dfrac{1}{802}-\dfrac{1}{810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{1}{10}-\dfrac{1}{810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}.\dfrac{8}{81}\)
\(\dfrac{1}{7}B=\dfrac{1.8}{8.81}\)
\(\dfrac{1}{7}B=\dfrac{1}{81}\)
\(B=\dfrac{1}{81}:\dfrac{1}{7}\)
\(B=\dfrac{7}{81}\)
\(A=\frac{1}{7}\left[\frac{1}{2}-\frac{1}{9}+...+\frac{1}{252}-\frac{1}{509}\right]\)
\(A=\frac{1}{7}.\left[\frac{1}{2}-\frac{1}{509}\right]\)
\(A=\frac{1}{7}.\left[\frac{507}{1018}\right]=\frac{507}{7126}\)
mk nghĩ là vậy đó, ủng hộ mk nha
Đặt \(A=\frac{1}{2.9}+\frac{1}{9.7}+\frac{1}{7.19}+...+\frac{1}{252.509}\)
\(\Leftrightarrow A=\frac{2}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{504.509}\right)\)
\(\Leftrightarrow A=\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{504}-\frac{1}{509}\right)\)
\(\Leftrightarrow A=\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{509}\right)\)
\(\Leftrightarrow A=\frac{2}{5}.\frac{505}{2036}\)
\(\Leftrightarrow A=\frac{101}{1018}\)
~ Hok tốt ~
#)Giải :
\(A=\frac{1}{2.9}+\frac{1}{9.7}+\frac{1}{7.19}+...+\frac{1}{252.509}\)
\(A=\frac{2}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{504.509}\right)\)
\(A=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{504}-\frac{1}{509}\right)\)
\(A=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{509}\right)\)
\(A=\frac{2}{5}\times\frac{505}{2036}\)
\(A=\frac{101}{1018}\)
\(A=\frac{2}{4.9}+\frac{2}{9.14}+\frac{2}{14.19}+...+\frac{2}{504.509}\)
\(A=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{504}-\frac{1}{509}\right)\)
\(A=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{509}\right)=...\)
\(B=\frac{1.4+1}{1.4}+\frac{4.7+1}{4.7}+\frac{7.10+1}{7.10}+...+\frac{100.103+1}{100.103}\)
\(B=1+\frac{1}{1.4}+1+\frac{1}{4.7}+...+1+\frac{1}{100.103}\)
\(B=34+\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(B=34+\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=34+\frac{1}{3}\left(1-\frac{1}{103}\right)=...\)
a)Ta có:
A= 1/2.9 + 1/9.7 +...+1/252.509
= 2/5.(5/4.9 + 5/9.14 + 5/14.19 +...+ 1/504.509)
= 2/5.(1/4 - 1/9 + 1/9 - 1/14 + 1/14 - 1/19 +...+ 1/504 - 1/509)
= 2/5.(1/4 - 1/509)
= 101/1018
Vậy A = 101/1018
b)Ta có:
B= 1/10.9 +1/18.13 + 1/26.17 +...+ 1/802.405)
= 1/4.(8/10.18 + 8/18.26 + 8/26.34 +...+ 8/802.810)
= 1/4.(1/10 - 1/18 + 1/18 - 1/26 + 1/26 - 1/34 +...+ 1/802 - 1/810)
= 1/4.(1/10 - 1/810)
= 2/81
Vậy B= 2/81
Tk mình nha!!!
\(\dfrac{5}{2}A=\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+...+\dfrac{5}{504.509}\)
\(\dfrac{5}{2}A=\dfrac{9-4}{4.9}+\dfrac{14-9}{9.14}+\dfrac{19-14}{14.19}+...+\dfrac{509-504}{504.509}\)
\(\dfrac{5}{2}A=\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{504}-\dfrac{1}{509}\)
\(\dfrac{5}{2}A=\dfrac{1}{4}-\dfrac{1}{509}\)
\(A=\left(\dfrac{1}{4}-\dfrac{1}{509}\right).\dfrac{2}{5}\)
\(A=\dfrac{1}{10}-\dfrac{2}{2545}< \dfrac{1}{10}\)
\(\Rightarrow A< \dfrac{1}{10}\)(đpcm)
Chúc bạn học tốt!
Ta có:
A=\(\dfrac{1}{2.9}+\dfrac{1}{9.7}+\dfrac{1}{7.19}+...+\dfrac{1}{252.509}\)
A=2.(\(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{504.509}\))
A=\(\dfrac{2}{5}\).(\(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{504}-\dfrac{1}{509}\))
A=\(\dfrac{2}{5}\).(\(\dfrac{1}{4}-\dfrac{1}{509}\))
A=\(\dfrac{2}{5}\).(\(\dfrac{509}{2036}-\dfrac{4}{2036}\))
A=\(\dfrac{2}{5}\).\(\dfrac{505}{2036}\)
A=\(\dfrac{101}{1018}\)