1/10x9+1/18x13+1/26x17+...+1/802x405
VT
Những câu hỏi liên quan
a/ 1/2x9 + 1/9x7+1/7x19+...........+1/252x509
b/1/10x9+1/18x13+1/26x17+..................+1/802x405
Ai BIẾT CÂU NÀY GIÚP MIK VỚI
a) \(\frac{1}{2.9}+\frac{1}{9.7}+\frac{1}{7.19}+...+\frac{1}{202.509}=\frac{2}{4.9}+\frac{2}{9.14}+\frac{2}{14.19}+...+\frac{2}{504.509}\)
\(=\frac{2}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{504.509}\right)\)
\(=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{504}-\frac{1}{509}\right)=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{509}\right)\)
\(=\frac{2}{5}.\frac{505}{2036}=\frac{101}{1018}\)
b) \(\frac{1}{10.9}+\frac{1}{18.13}+...+\frac{1}{802.405}=\frac{2}{10.18}+\frac{2}{18.26}+...+\frac{2}{802.810}\)
\(=\frac{2}{8}\left(\frac{8}{10.18}+\frac{8}{18.26}+...+\frac{8}{802.810}\right)=\frac{1}{4}\left(\frac{1}{10}-\frac{1}{18}+\frac{1}{18}-\frac{1}{26}+...+\frac{1}{802}-\frac{1}{810}\right)\)
\(=\frac{1}{4}\left(\frac{1}{10}-\frac{1}{810}\right)=\frac{1}{4}.\frac{40}{405}=\frac{10}{405}\)
Bạn vào câu hỏi tương tự tham khảo !
a) C=1/9x10+1/18x13+1/26x17+.......+1/802x405
b)D=2/4x7-3/5x9+2/7x10-3/9x13+......+2/302x304+3/401x405
Qfrac{1}{2x9}+frac{1}{9x7}+frac{1}{7x19}+.......+frac{1}{252x509}Rfrac{1}{10x9}+frac{1}{18x13}+frac{1}{26x17}+.....+frac{1}{802x405}Sfrac{2}{4x7}-frac{3}{5x9}+frac{2}{7x10}-frac{3}{9x13}+.........+frac{2}{301x304}+frac{3}{401x403}
Đọc tiếp
Q=\(\frac{1}{2x9}\)+\(\frac{1}{9x7}\)+\(\frac{1}{7x19}\)+.......+\(\frac{1}{252x509}\)
R=\(\frac{1}{10x9}\)+\(\frac{1}{18x13}\)+\(\frac{1}{26x17}\)+.....+\(\frac{1}{802x405}\)
S=\(\frac{2}{4x7}\)\(-\)\(\frac{3}{5x9}\)+\(\frac{2}{7x10}\)\(-\)\(\frac{3}{9x13}\)+.........+\(\frac{2}{301x304}\)+\(\frac{3}{401x403}\)
Q=\(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{7}+\frac{1}{7}-\frac{1}{19}+...+\frac{1}{252}-\frac{1}{509}\)
=\(\frac{1}{2}-\left(\frac{1}{9}+\frac{1}{9}\right)-\left(\frac{1}{7}+\frac{1}{7}\right)-...-\left(\frac{1}{252}+\frac{1}{252}\right)-\frac{1}{509}\)
=\(\frac{1}{2}-0+0+0+...+0-\frac{1}{509}\)
=\(\frac{1}{2}-\frac{1}{509}\)
=\(\frac{507}{1018}\)
MẤY CÂU KHÁC THÌ TƯƠNG TỰ, CHÚC BẠN MAY MẮN!!!:))
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làm 2 câu còn lại đi câu đó làm rồi
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Tìm A= 9^2018-10x9^2017+10x9^2016-...+10x9^2-10x9+1
giúp mình nhé ,đg gấp
Tìm A=9^2018-10x9^2017+10x9^2016-...+10x9^2-10x9+1
giúp mình nhé,đg gấp
TÍNH TỔNG S=1/5x6+1/10x9+1/15x12+.....+1/3350x2013
TÍNH TỔNG S=1/5x6+1/10x9+1/15x12+.....+1/3350x2013
tính A=9^2017-10x9^2016+10x9^2015-10x9^2014+....+10x9^3-10x9^2+10x9-10
A=288-18x13+567:17
B=\(\frac{1}{7}\)x\(3\frac{1}{11}\)\(-\frac{9}{11}\)\(+\frac{1}{7}\)
A = 288 - 18x13 +567:17
A = 288 - 234 + \(\frac{567}{17}\)
A = \(\frac{1485}{17}\)
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\(A=288-18\cdot13+567:17\)
\(A=288-234+567:17\)
\(A=288-234+\frac{567}{17}\)
\(A=54+\frac{567}{17}\)
\(A=87\frac{6}{17}=\frac{1485}{17}\)
\(B=\frac{1}{7}\cdot3\frac{1}{11}-\frac{9}{11}+\frac{1}{7}\)
\(B=\frac{1}{7}\cdot\frac{34}{11}-\frac{9}{11}+\frac{1}{7}\)
\(B=\frac{34}{77}-\frac{9}{11}+\frac{1}{7}\)
\(B=\frac{-29}{77}+\frac{1}{7}\)
\(B=\frac{-18}{77}\)
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A = 288 - 18 x 13 + 567 : 17
A = 288 - 234 + 567 : 17
A = 288 - 234 + \(\frac{567}{17}\)
A = 54 + \(\frac{567}{17}\)
A = 87\(\frac{6}{17}\)= \(\frac{1485}{17}\)
B = \(\frac{1}{7}\)x 3\(\frac{1}{11}\) - \(\frac{9}{11}\) + \(\frac{1}{7}\)
B = \(\frac{1}{7}\)x 3\(\frac{1}{11}\)-\(\frac{9}{11}\) + \(\frac{1}{7}\)
B = \(\frac{34}{77}\) - \(\frac{9}{11}\) + \(\frac{1}{7}\)
B = \(\frac{-29}{77}\) + \(\frac{1}{7}\)
B = \(\frac{-18}{77}\)
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