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S
3
5 tháng 3 2020
A=-4/4.5+-4/5.6+-4/6.7+-4/7.8+-4/8.9+-4/9.10
-1/4A=1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
-1/4A=1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10
-1/4A=1/4-1/10
-1/4A=3/20
A=-3/5
S
1
5 tháng 3 2020
\(A=\frac{-4}{20}+\frac{-4}{30}+\frac{-4}{42}+...+\frac{-4}{90}\)
=\(-4\left(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}\right)\)
=-4\(\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{9.10}\right)\)
=-4\(\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{10}\right)\)
=-4\(\left(\frac{1}{4}-\frac{1}{10}\right)\)
=-4.\(\frac{3}{20}\)=\(\frac{-3}{5}\)
S
0
Q
11 tháng 4 2017
\(\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)
\(=\dfrac{-1}{4.5}+\dfrac{-1}{5.6}+\dfrac{-1}{6.7}+\dfrac{-1}{7.8}+\dfrac{-1}{8.9}+\dfrac{-1}{9.10}\)
\(=\dfrac{-1}{4}-\dfrac{-1}{5}+\dfrac{-1}{5}-\dfrac{-1}{6}+\dfrac{-1}{6}-\dfrac{-1}{7}+\dfrac{-1}{7}-\dfrac{-1}{8}+\dfrac{-1}{8}-\dfrac{-1}{9}+\dfrac{-1}{9}-\dfrac{-1}{10}\)
\(=\dfrac{-1}{4}-\dfrac{-1}{10}\)
\(=\dfrac{-3}{20}\)
NM
3
9 tháng 9 2020
a)A=1/20+1/30+1/42+1/56+1/72+1/90+1/110
= 1/4*5 + 1/5*6 + 1/6*7 + ... + 1/10*11
= 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/10 - 1/11
= 1/4 - 1/11
= 7/44
b)B=1/2+1/4+1/6+1/8+...+1/512+1/1024
B = 1/2^1 + 1/2^2 + 1/2^3 + ... + 1/2^9 + 1/2^10
2B = 1 + 1/2 + 1/2^2 + ... + 1/2^10 + 1/2^11
2B - B = B = 1 + 1/2^11
9 tháng 3 2018
a. \(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+......+\dfrac{3}{17.20}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+......+\dfrac{1}{17}-\dfrac{1}{20}\)
\(=\dfrac{1}{2}-\dfrac{1}{20}\)
\(=\dfrac{9}{20}\)
b. \(B=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{4}-\dfrac{1}{10}\)
\(=\dfrac{3}{20}\)
c. \(C=\dfrac{4^2}{1.5}+\dfrac{4^2}{5.9}+......+\dfrac{4^2}{45.49}\)
\(=4\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+....+\dfrac{4}{45.49}\right)\)
\(=4\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+.....+\dfrac{1}{45}-\dfrac{1}{49}\right)\)
\(=4\left(1-\dfrac{1}{49}\right)\)
\(=4.\dfrac{48}{49}\)
\(=\dfrac{192}{49}\)
Ta có: \(A=-\frac{4}{20}-\frac{4}{30}-\frac{4}{42}-\frac{4}{56}-\frac{4}{72}-\frac{4}{90}\)
\(\Leftrightarrow A=-4.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(\Leftrightarrow A=-4.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(\Leftrightarrow A=-4.\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(\Leftrightarrow A=-4.\left(\frac{5-2}{20}\right)\)
\(\Leftrightarrow A=-4.\frac{3}{20}=-\frac{3}{5}\)
Vậy \(\Leftrightarrow A=-\frac{3}{5}\)
thanks