Tính C : ( 4/1x3 + 4/3x5 + 4/x7 + ... + 4/97x99 ) - 5/11x12 - 5/12x13 - 5/13x14 -...- 5/98x99
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HT
5
25 tháng 6 2017
\(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+...+\frac{5}{97\cdot99}=\frac{5}{2}\left[\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{97\cdot99}\right]\)
\(=\frac{5}{2}\left[\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=\frac{5}{2}\left[1-\frac{1}{99}\right]\)
\(=\frac{5}{2}\cdot\frac{98}{99}=\frac{245}{99}\)
25 tháng 6 2017
\(=\frac{5}{2}\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{97\times99}\right)\)
\(=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{5}{2}\left(1-\frac{1}{99}\right)\)
\(=\frac{5}{2}\times\frac{98}{99}\)
\(=\frac{245}{99}\)
HT
4
30 tháng 6 2017
\(\frac{5}{11x12}+\frac{5}{12x13}+...+\frac{5}{98x99}\)
=\(\frac{5}{11}-\frac{5}{12}+\frac{5}{12}-\frac{5}{13}+...+\frac{5}{98}-\frac{5}{99}\)
=\(\frac{5}{11}-\frac{5}{99}\)
=\(\frac{40}{99}\)
Q
30 tháng 6 2017
Cái cuối bỏ 1 số 0 thì đúng hơn nha bạn
\(\frac{5}{11.12}+\frac{5}{12.13}+\frac{5}{13.14}+...+\frac{5}{98.99}\)
\(=5\left(\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+...+\frac{1}{98.99}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{99}\right)\)
\(=5.\frac{8}{99}\)
\(=\frac{40}{99}\)
CC
7
PL
1 tháng 4 2019
\(E=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)
\(\Rightarrow E=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\) (đặt 2 làm nhân tử chung để ta có các số hạng trong ngoặc có hiệu 2 số ở mẫu = tử)
\(\Rightarrow E=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow E=2.\left(1-\frac{1}{99}\right)\)
\(\Rightarrow E=2.\frac{98}{99}\)
\(\Rightarrow E=\frac{196}{99}\)
*Không biết có đúng ko :)
16 tháng 8 2019
\(a,x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x-\frac{61}{8}=\frac{5}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{10}{8}+\frac{61}{8}=\frac{71}{8}=8\frac{7}{8}\)
\(b,x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x+\frac{43}{5}=\frac{37}{4}\)
=> \(x=\frac{37}{4}-\frac{43}{5}=\frac{13}{20}\)
\(c,\left[x-7\frac{5}{8}\right]:\frac{1}{2}=3\)
=> \(\left[x-\frac{61}{8}\right]=3\cdot\frac{1}{2}\)
=> \(\left[x-\frac{61}{8}\right]=\frac{3}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}=\frac{12}{8}+\frac{61}{8}=\frac{73}{8}=9\frac{1}{8}\)
d, \(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+...+\frac{x}{97\cdot99}=99\)
=> \(\frac{x}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\cdot\frac{98}{99}=99\)
=> \(\frac{98x}{198}=99\)
=> 98x = 99 . 198
=> 98x = 19602
=> x = 19602 : 98 = 9801/49
16 tháng 8 2019
a) \(x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{71}{8}\)
b) \(x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x=\frac{37}{4}-\frac{61}{8}\)
=> \(x=\frac{13}{8}\)
c) \(\left(x-7\frac{5}{8}\right):\frac{1}{2}=3\)
=> \(x-\frac{61}{8}=3.\frac{1}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}\)
=> \(x=\frac{73}{8}\)
d) \(\frac{x}{1.3}+\frac{x}{3.5}+...+\frac{x}{97.99}=99\)
=> \(x.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)=99\)
=> \(\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)=99\)
=> \(x\left(1-\frac{1}{99}\right)=99:\frac{1}{2}\)
=> \(x.\frac{98}{99}=198\)
=> \(x=198:\frac{98}{99}=\frac{9801}{49}\)
S
5
NH
0
ML
24 tháng 6 2017
\(M=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)
\(M=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(M=2\left(1-\frac{1}{100}\right)\)
\(M=2.\frac{99}{100}\)
\(M=\frac{99}{50}\)
\(N=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{97.99}\)
\(N=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(N=\frac{3}{2}\left(1-\frac{1}{99}\right)\)
\(N=\frac{3}{2}.\frac{98}{99}\)
\(N=\frac{49}{33}\)
9 tháng 7 2018
= \(\frac{1}{2}\)- \(\frac{2}{3}\)+ (\(\frac{3}{4}\)- \(\frac{3}{4}\)) + ( -\(\frac{4}{5}\)+ \(\frac{4}{5}\)) + ( \(\frac{5}{6}-\frac{5}{6}\)) - \(\frac{6}{7}\)
= \(\frac{1}{2}-\frac{2}{3}-0-0-0-\frac{6}{7}\)
= \(\frac{1}{2}-\frac{2}{3}-\frac{6}{7}\)
=\(\frac{21}{42}-\frac{28}{42}-\frac{36}{42}\)
= \(\frac{-43}{42}\)
