\(B=\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)
các bạn nhớ trình bày bài giải đầy đủ nhé mình tích cho.
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MH
21 tháng 2 2016
7/1.3 + 7/3.5 + 7/5.7 + ... + 7/99.101
= 7.(1/1.3 + 1/3.5 + 1/5.7 + ... + 1/99.101)
= 7/2 . 2 . (1/1.3 + 1/3.5 + 1/5.7 + ... + 1/99.101)
= 7/2 . (2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101)
= 7/2 . (1 - 1/3 + 1/3 - 1/5 + ... + 1/99 - 1/101)
= 7/2 . (1 - 1/101)
= 7/2 . 100/101
= 350/101
NK
21 tháng 2 2016
\(\frac{7}{1.3}+\frac{7}{3.5}+...+\frac{7}{99.101}\)
\(=7\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)
\(=\)\(\frac{7}{2}.2.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)
\(=\)\(\frac{7}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
VT
2
7 tháng 3 2017
\(\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+....+\frac{7}{99.101}\)
\(=\frac{7}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\right)\)
\(=\frac{7}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}\left(1-\frac{1}{101}\right)=\frac{7}{2}.\frac{100}{101}=\frac{350}{101}\)
DT
2
N
25 tháng 3 2017
\(\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)
\(=\frac{7}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{7}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\frac{100}{101}\)
\(=\frac{350}{101}\)
k mk nha
25 tháng 3 2017
7/1.3+7/3.5+7/5.7+...+7/99.101
=7(1/1.3+1/3.5+1/5.7+...+1/99.101)
=7(1/1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)
=7(1-1/101)
=7.100/101
=700/101
Đầy đủ ko bỏ bước nào lun!!
K CHO MK NHA!!!
PT
8
19 tháng 7 2018
=1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
=1-1/101
=100/101
k cho mình nha
19 tháng 7 2018
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{101}\right)=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)
DD
Đoàn Đức Hà
Giáo viên
18 tháng 5 2021
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(B=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(B=1-\frac{1}{101}=\frac{100}{101}\)
DD
Đoàn Đức Hà
Giáo viên
18 tháng 5 2021
\(C=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)
\(C=3\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\right)\)
\(C=3\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{20-17}{17.20}\right)\)
\(C=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(C=3\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{27}{20}\)
\(D=\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)
\(D=\frac{7}{2}B=\frac{7}{2}.\frac{100}{101}=\frac{350}{101}\)
29 tháng 6 2016
\(A=\frac{7}{1.3}+\frac{7}{3.5}+.............+\frac{7}{99.101}\)
\(=\frac{7}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+........+\frac{2}{99.101}\right)\)
\(=\frac{7}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.......+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\frac{100}{101}\)
\(=\frac{350}{101}\)
Q
16 tháng 4 2017
\(\frac{202}{1.3}+\frac{202}{3.5}+\frac{202}{5.7}+...+\frac{202}{99.101}\)
\(=202\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(=202.\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=101\left(1-\frac{1}{101}\right)\)
\(=101.\frac{100}{101}\)
\(=100=10^2\Rightarrowđpcm\)
M
16 tháng 1 2016
a) \(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}+\frac{3}{418}+\frac{3}{550}\)
= \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}+\frac{3}{19.22}+\frac{3}{22.25}\)
= \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+\frac{1}{19}-\frac{1}{22}+\frac{1}{22}-\frac{1}{25}\)
= \(\frac{1}{1}-\frac{1}{25}\)
= \(\frac{24}{25}\)
b) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n+1\right).\left(2n+3\right)}\)
= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n+1}-\frac{1}{2n+3}\)
= \(\frac{1}{1}-\frac{1}{2n+3}\)
= \(\frac{2n+2}{2n+3}\)
c) \(\frac{7+\frac{7}{13}-\frac{7}{48}+\frac{7}{95}}{15+\frac{15}{13}-\frac{15}{48}+\frac{15}{95}}-\frac{7070707}{15151515}\)
= \(\frac{7\left(1+\frac{1}{13}-\frac{1}{48}+\frac{1}{95}\right)}{15\left(1+\frac{1}{13}-\frac{1}{48}+\frac{1}{95}\right)}-\frac{7.1010101}{15.1010101}\)
= \(\frac{7}{15}-\frac{7}{15}\)
= 0
B : 7/2 =2/1.3+2/3.5+...+2/99.101
B:7/2=1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
B:7/2=1-1/101=100/101
B=100/101*7/2=700/202=350/101
B=7/2(2/1.3+2/3.5+ ...+2/99.101)
B=7/2(1-1/3+1/3-1/5+...+1/99-1/101)
B=7/2(1-1/101)=7/2.100/101=350/101
k nha bạn