tính nhanh
A = 1x3 + 3x5 + 5x7 + .............+ 37x39 + 39x41
K
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Những câu hỏi liên quan
Q3
2
VD
11 tháng 3 2022
\(A=\dfrac{1}{3.5}+\dfrac{1}{7.9}+...+\dfrac{1}{37.39}\\ =\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{7.9}+...+\dfrac{2}{37.39}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{39}\right)\\ =\dfrac{1}{2}.\dfrac{4}{13}\\ =\dfrac{2}{13}\)
HT
27 tháng 10 2020
sửa đề câu a và câu b nhá , mik nghĩ đề như này :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
= \(\frac{1}{1}-\frac{1}{215}\)
\(=\frac{214}{215}\)
b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)
\(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)
\(A\cdot2=\frac{214}{215}\)
\(A=\frac{214}{215}:2\)
\(A=\frac{107}{215}\)
LT
2
VD
13 tháng 3 2022
\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+...+\dfrac{2}{99\times101}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =1-\dfrac{1}{101}\\ =\dfrac{100}{101}\)
TH
2
TM
24 tháng 6 2018
a) \(a=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{37\cdot39}\)
\(a=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)
\(a=\frac{1}{3}-\frac{1}{39}\)
\(a=\frac{12}{39}\)
b) \(\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)
\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot\left(\frac{-5}{12}+\frac{3}{12}+\frac{2}{12}\right)\)
\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot\left(\frac{-2}{12}+\frac{2}{12}\right)\)
\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot0\)
\(=0\)
24 tháng 6 2018
a) \(A=\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{37x39}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{37}-\frac{1}{39}\)
\(A=\frac{1}{3}-\frac{1}{39}\)
\(A=\frac{4}{13}\)
b) \(\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)x\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)
\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)x0\)
\(=0\)
SL
0
1 tháng 5 2019
1/2 x (6/1-6/3+6/3-6/5+ ... +6/37-6/39)
1/2 x (6/1-6/39)
1/2 x 228/39
228/78
DD
6 tháng 8 2015
\(\frac{3}{1x3}+\frac{3}{3x5}+...+\frac{3}{49x51}=\frac{3}{2}\left(\frac{2}{1x3}+\frac{2}{3x5}+...+\frac{2}{49x51}\right)=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)
NK
1
11 tháng 5 2015
\(\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{19\times21}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\)
\(=1-\frac{1}{21}=\frac{20}{21}\)
đúng cái nhé
PN
3
2 tháng 7 2018
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{2017.2019}\)
\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{3}{2}.\left(1-\frac{1}{2019}\right)\)
\(=\frac{3}{2}.\frac{2018}{2019}\)
\(=\frac{1009}{673}\)
2 tháng 7 2018
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}.....+\frac{3}{2017.2019}\)
\(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{2017.2019}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{3}+....+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{2019}\right)\)
\(=\frac{3}{2}.\frac{2018}{2019}=\frac{1009}{673}\)
Ta có : A = 1.3 + 3.5 + 5.7 + ....... + 37.39 + 39.41
=> 6A = 1.3.5 - 1.3.5 + 3.5.7 - 3.5.7 + ....... + 39.41.43
=> 6A = 39.41.43
=> A = 39.41.43 / 6
=> A = 11459,5
A=1x3+3x5+5x7+...+37x39+39x41
6A=1x3x5-1x3x5+3x5x7-3x5x7+...+39x41x43
6A=39x41x43
A=\(\frac{39x41x43}{6}\)