C=2/4.7-3/5.9+2/7.10-3/9.13+...+3/301.304-3/401.405
A=\(\dfrac{2}{4.7}\)\(-\dfrac{3}{5.9}\)\(+\dfrac{2}{7.10}\)\(-\dfrac{3}{9.13}\)\(+.....\)\(+\dfrac{2}{301.304}\)\(-\dfrac{3}{401.405}\)
tính A
\(A=\dfrac{2}{4.7}-\dfrac{3}{5.9}+\dfrac{2}{7.10}-\dfrac{3}{9.13}+...+\dfrac{2}{301.304}-\dfrac{3}{401.405}\)
\(A=\dfrac{2}{4.7}+\dfrac{2}{7.10}+\dfrac{2}{301.304}...-\left(\dfrac{3}{5.9}+\dfrac{3}{9.13}+...+\dfrac{3}{401.405}\right)\)
\(A=2\left(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{301.304}\right)...-3\left(\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{401.405}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{301}-\dfrac{1}{304}\right)...-3\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{401}-\dfrac{1}{405}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{304}\right)-3\left(\dfrac{1}{5}-\dfrac{1}{405}\right)\)
\(A=2\left(\dfrac{76}{304}-\dfrac{1}{304}\right)-3\left(\dfrac{81}{5}-\dfrac{1}{405}\right)\)
\(A=2.\dfrac{75}{304}-3.\dfrac{80}{405}=\dfrac{75}{152}-\dfrac{80}{135}=\dfrac{10125-12160}{152.135}=-\dfrac{2035}{152.135}=-\dfrac{407}{4104}\)
C = \(\frac{2}{4.7}-\frac{3}{5.9}+\frac{2}{7.10}-\frac{3}{9.13}+...+\frac{2}{301.304}-\frac{3}{401.405}\)
tính c
TẬP HỢP RA HAI NHÓM .MỘT NHÓM SỐ ÂM.CÒN NHÓM KIA LÀ SỐ DƯƠNG MÀ TÍNH
STUDY WELL
K NHA
MK XIN CẢM ƠN CÁC BẠN NHÌU
C = 24.7 −35.9 +27.10 −39.13 +...+2301.304 −3401.405
\(C=\left(\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{301.304}\right)-\left(\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{401.405}\right)\)
\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{401}-\frac{1}{405}\right)\)
\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{405}\right)\)
\(C=\frac{2}{3}.\frac{75}{304}-\frac{3}{4}.\frac{16}{81}\)
\(C=\frac{25}{152}-\frac{4}{27}\)
\(C=\frac{67}{4104}\)
Study well
\(C=\frac{2}{4\cdot7}-\frac{3}{5\cdot9}+\frac{2}{7\cdot10}-\frac{3}{9\cdot13}+...+\frac{2}{301\cdot304}-\frac{3}{401\cdot405}\)
\(C=\left(\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{301\cdot304}\right)-\left(\frac{3}{5\cdot9}+\frac{3}{9\cdot13}+...+\frac{3}{401\cdot405}\right)\)
\(C=\frac{2}{3}\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{301\cdot304}\right)-\frac{3}{4}\left(\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{401\cdot405}\right)\)
\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{401}-\frac{1}{405}\right)\)
\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{405}\right)\)
\(C=\frac{2}{3}\cdot\frac{77}{304}-\frac{3}{4}\cdot\frac{82}{405}\)
\(C=\frac{77}{456}-\frac{41}{270}\)
\(C=\frac{349}{20520}\)
Không chắc =))
Tính :
a)A=1/2.9 + 1/9.7 + 1/7.19 + ...+1/252.509
b)B=1/10.9 + 1/18.13 + 1/26.17 + ... + 1/802.405
c)C=2/4.7 - 3/5.9 +2/7.10 - 3/9.13+...+2/301.304 - 3/401.405
\(A=\frac{1}{2.9}+\frac{1}{9.7}+...+\frac{1}{252.509}\)
\(A=\frac{2}{4.9}+\frac{2}{9.14}+...+\frac{2}{504.509}\)
\(A=\frac{2}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+...+\frac{5}{504.509}\right)\)
\(A=\frac{2}{5}.\left(\frac{9-4}{4.9}+\frac{14.9}{9.14}+...+\frac{509-504}{504.509}\right)\)
\(A=\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{504}-\frac{1}{509}\right)\)
\(A=\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{509}\right)\)
\(A=\frac{2}{5}.\frac{505}{2036}\)
\(A=\frac{101}{1018}\)
\(B=\frac{1}{10.9}+\frac{1}{18.13}+\frac{1}{26.17}+...+\frac{1}{802.405}\)
\(\frac{1}{2}B=\frac{1}{10.9.2}+\frac{1}{18.13}+\frac{1}{26.17}+...+\frac{1}{802.405.2}\)
\(\frac{1}{2}B=\frac{1}{10.18}+\frac{1}{18.26}+\frac{1}{26.34}+...+\frac{1}{802.810}\)
\(4B=\frac{8}{10.18}+\frac{8}{18.26}+\frac{8}{26.34}+...+\frac{8}{802.810}\)
\(4B=\frac{18-10}{10.18}+\frac{26-18}{28.26}+\frac{34-26}{26.34}+...+\frac{810-802}{802.810}\)
\(4B=\frac{1}{10}-\frac{1}{18}+\frac{1}{18}-\frac{1}{26}+\frac{1}{26}-\frac{1}{34}+...+\frac{1}{802}-\frac{1}{810}\)
\(4B=\frac{1}{10}-\frac{1}{810}\)
\(4B=\frac{8}{81}\)
\(B=\frac{2}{81}\)
\(C=\frac{2}{4.7}-\frac{3}{5.9}+\frac{2}{7.10}-\frac{3}{9.13}+...+\frac{2}{301.304}-\frac{3}{401.405}\)
\(C=\left(\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{301.304}\right)-\left(\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{401.405}\right)\)
\(C=\frac{2}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right)-\frac{3}{4}.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{401}-\frac{1}{405}\right)\)
\(C=\frac{2}{3}.\left(\frac{1}{4}-\frac{1}{304}\right)-\frac{3}{4}.\left(\frac{1}{5}-\frac{1}{405}\right)\)
\(C=\frac{2}{3}.\frac{75}{304}-\frac{3}{4}.\frac{16}{81}\)
\(C=\frac{25}{152}-\frac{4}{27}\)
\(C=\frac{67}{4104}\)
C=\(\frac{2}{4.7}-\frac{3}{5.9}+\frac{2}{7.10}-\frac{3}{9.13}+...........+\frac{2}{301.304}-\frac{3}{401.405}\)
Các bạn ơi trả lời giúp mình nhanh nhanh nha!
Ai trả lời đúng mình sẽ tick
\(C=\frac{2}{4.7}-\frac{3}{5.9}+\frac{2}{7.10}-\frac{3}{9.13}+...+\frac{2}{301.304}-\frac{3}{401.405}\)
\(C=\left(\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{301.304}\right)-\left(\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{401.405}\right)\)
\(C=\frac{2}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{301.304}\right)-\frac{3}{4}\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{401.405}\right)\)
\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+..+\frac{1}{401}-\frac{1}{405}\right)\) \(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{405}\right)\)
\(C=\frac{25}{152}-\frac{4}{27}\)
\(C=\frac{67}{4104}\)
tính
\(\frac{2}{4.7}-\frac{3}{5.9}+\frac{2}{7.10}-\frac{3}{9.12}+...+\frac{2}{301.304}-\frac{3}{401.405}\)
Tính :
c) C = \(\dfrac{2}{4.7}-\dfrac{3}{5.9}+\dfrac{2}{7.10}-\dfrac{3}{9.13}+...+\dfrac{2}{301.304}-\dfrac{3}{401.405}\)
d) D = \(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\left(\dfrac{1-3-5-7-...-49}{89}\right)\)
\(\frac{1}{10.9}+\frac{1}{18.13}+\frac{1}{26.17}+...+\frac{1}{802.405}\)
2.
\(\frac{2}{4.7}-\frac{3}{5.9}+\frac{2}{7.10}-\frac{3}{9.13}+.......+\frac{2}{301.304}-\frac{3}{401.405}\)
ai tính nhành và đúng nhất tớ sẽ tick cho người đó
Tính:
\(\frac{2}{4.7}\) - \(\frac{3}{5.9}\) + \(\frac{2}{7.10}\) - \(\frac{3}{9.13}\) + ... + \(\frac{2}{301.304}\) - \(\frac{3}{401.405}\)
Sử dụng điều kiện \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) để tính:
a. \(A=\dfrac{1}{2.9}+\dfrac{1}{9.7}+\dfrac{1}{7.19}+...+\dfrac{1}{252.509}\)
b.\(B=\dfrac{2}{4.7}-\dfrac{3}{5.9}+\dfrac{2}{7.10}-\dfrac{3}{9.13}+...+\dfrac{2}{301.304}-\dfrac{3}{401.405}\)
A=1/2.9+1/9.7+1/7.19+...+1/252.509
=?
??????