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D
17 tháng 4 2017
\(\frac{3}{5.6}+\frac{3}{6.7}+......+\frac{3}{11.12}=\frac{1}{6}X\)
\(\Rightarrow3.\left(\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{11.12}\right)=\frac{1}{6}X\)
\(\Rightarrow3.\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+.....+\frac{1}{11}-\frac{1}{12}\right)=\frac{1}{6}X\)
\(\Rightarrow3.\left(\frac{1}{5}-\frac{1}{12}\right)=\frac{1}{6}X\)
\(\Rightarrow3.\frac{7}{60}=\frac{1}{6}X\)
\(\Rightarrow\frac{21}{60}=\frac{1}{6}X\)
\(\Rightarrow X=\frac{21}{60}\div\frac{1}{6}=\frac{21}{10}\)
Vậy \(X=\frac{21}{10}\)
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. \(C=\dfrac{4^2}{1.5}+\dfrac{4^2}{5.9}+......+\dfrac{4^2}{45.49}\)
\(=4\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+....+\dfrac{4}{45.49}\right)\)
\(=4\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+.....+\dfrac{1}{45}-\dfrac{1}{49}\right)\)
\(=4\left(1-\dfrac{1}{49}\right)\)
\(=4.\dfrac{48}{49}\)
\(=\dfrac{192}{49}\)
LH
1
14 tháng 7 2015
A - B = \(\left(1+\frac{1}{2}+1+\frac{1}{12}+1+\frac{1}{30}+1+\frac{1}{56}+1+\frac{1}{90}\right)-\left(1-\frac{1}{6}+1-\frac{1}{20}+1-\frac{1}{42}+1-\frac{1}{72}+1-\frac{1}{110}\right)\)= \(\left(5+\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}\right)-\left(5-\frac{1}{6}-\frac{1}{20}-\frac{1}{42}-\frac{1}{72}-\frac{1}{110}\right)\)\
= \(5+\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}-5+\frac{1}{6}+\frac{1}{20}+\frac{1}{42}+\frac{1}{72}+\frac{1}{110}\)
= \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}+\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{10.11}\)
= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}=1-\frac{1}{11}=\frac{10}{11}\)
7 tháng 2 2016
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+....+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)=9-\frac{9}{10}=\frac{81}{10}\)
7 tháng 2 2016
b.=3/2.4/3....2012/2011
=3.4....2012/2.3....2011=2012/2=1006
BT
0
Answer:
\(\frac{3}{20}+\frac{3}{30}+\frac{3}{42}+\frac{3}{56}+\frac{3}{72}+\frac{3}{90}\)
\(=\frac{3}{20}+\frac{1}{10}+\frac{1}{14}+\frac{3}{56}+\frac{1}{24}+\frac{1}{30}\)
\(=\frac{126}{840}+\frac{84}{840}+\frac{60}{840}+\frac{45}{840}+\frac{35}{840}+\frac{28}{840}\)
\(=\frac{378}{840}=\frac{9}{20}\)