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NH
3
NQ
0
NT
1
15 tháng 7 2018
Đặt A = 1.3 + 3.5 + 5.7 + ... + 15.17 + 17.19
6A = 1.3.6 + 3.5.6 + .... + 15.17.6 + 17.19.6
6A = 1.3.(5 + 1) + 3.5.(7 - 1) + ... + 15.17.(19 - 13) + 17.19.(21 - 15)
6A = 1.3.5 + 1.3 + 3.5.7 - 1.3.5 + ... + 15.17.19 - 13.15.17 + 17.19.21 - 15.17.19
6A = (1.3.5 + 3.5.7 + ... + 15.17.19 + 17.19.21) - ( 1.3.5 + ... + 13.15.17 + 15.17.19) + 1.3
6A = 17.19.21 + 1.3
6A = 6783 + 3
6A = 6786
A = 6780 : 6 = 1131
VD
11 tháng 9 2017
chu so tan cung o ve phai la 0 vi co so 10 tan cung la 0. ve trai tan cung la 5 vi 5nhan so le nao cung tan cung la 5 vay chu so tan cung la 0-5 =5
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 1 2024
Bài 1:
a; \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{7}{21}\) + (- \(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{1}{3}\) -\(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{10}{36}\)) + (\(\dfrac{8}{19}\) + \(\dfrac{11}{19}\)) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{5}{18}\)) + \(\dfrac{19}{19}\) - 0 - \(\dfrac{5}{8}\)
= 0 + 1 - \(\dfrac{5}{8}\)
= \(\dfrac{3}{8}\)
b; \(\dfrac{1}{13}\) + (\(\dfrac{-5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\)) - (\(\dfrac{12}{17}\) - \(\dfrac{5}{18}\) + \(\dfrac{7}{5}\))
= \(\dfrac{1}{13}\) - \(\dfrac{5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{12}{17}\) + \(\dfrac{5}{18}\) - \(\dfrac{7}{5}\)
= (\(\dfrac{1}{13}\) - \(\dfrac{1}{13}\)) + (\(\dfrac{12}{17}\) - \(\dfrac{12}{17}\)) + (-\(\dfrac{5}{18}\) + \(\dfrac{5}{18}\)) - \(\dfrac{7}{5}\)
= 0 + 0 + 0 - \(\dfrac{7}{5}\)
= - \(\dfrac{7}{5}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 1 2024
Bài 1 c;
\(\dfrac{15}{14}\) - (\(\dfrac{17}{23}\) - \(\dfrac{80}{87}\) + \(\dfrac{5}{4}\)) + (\(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\))
= \(\dfrac{15}{14}\) - \(\dfrac{17}{23}\) + \(\dfrac{80}{87}\) - \(\dfrac{5}{4}\) + \(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\)
= (\(\dfrac{15}{14}-\dfrac{15}{14}\)) + (\(-\dfrac{17}{23}+\dfrac{17}{23}\)) - (\(\dfrac{5}{4}\) - \(\dfrac{1}{4}\)) + \(\dfrac{80}{87}\)
= 0 + 0 - 1 + \(\dfrac{80}{87}\)
= - \(\dfrac{7}{87}\)
N
2
5 tháng 3 2020
=1/1.3.5+1/3/5/7+1/5.7.9+......+1/17/19/21
=1/4.(5-1/1.3.5+7-3/3.5.7+.....+21-17/17/19/21
=1/4.(5/1.3.5-1/1.3.5+7/3.5.7-3/3.5.7+.....+21/17.19.21-17/17.19.21
=1/4.(1/1.3-1/3.5+1/3.5-1/5.7+.....+1/17.19-1/19.21)
=1/4.(1/3.1/21.17)
=1/4.3200/9603
= 800/9603
Chúc bạn học tốt^^
C
5 tháng 3 2020
Đặt \(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{17.19.21}\)
\(\Rightarrow4A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{17.19.21}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{17.19}-\frac{1}{19.21}\)
\(=\frac{1}{1.3}-\frac{1}{19.21}=\frac{44}{133}\)
\(\Rightarrow A=\frac{44}{133}\div4=\frac{11}{133}\)
29 tháng 4 2019
A = ( -4/5 + 4/3 ) + (-5/4 + 14/5) - 7/3
= 8/15 + 31/20 - 7/3
= 25/12 - 7/3
= -1/4
B = 8/3 x 2/5 x 3/8 x 10x 19/92
= 16/15 x 15/4 x 19/92
= 4x19/92
= 19/23
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
22 tháng 9 2024
C = - \(\dfrac{5}{7}\) x \(\dfrac{2}{11}\) + \(\dfrac{-5}{7}\) x \(\dfrac{9}{14}\) + \(\dfrac{1}{57}\)
= - \(\dfrac{10}{77}\) - \(\dfrac{45}{98}\) + \(\dfrac{1}{57}\)
= - \(\dfrac{635}{1078}\) + \(\dfrac{1}{57}\)
= - \(\dfrac{36195}{61446}\) + \(\dfrac{1078}{61446}\)
= - \(\dfrac{35117}{61446}\)
1+-3+5+-7+....+17+-19
=[1+(-3)]+[5+(-7)]+...+[17+(-19)]
= (-2)+(-2)+...+(-2) (co 5 so -2)
=-2.5=-10
moi nguoi an dung nha
đáp án = -10