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ta nhân 3 cả hai vế, được :
\(\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{102.105}\right)x=3\)
hay
\(\left(\frac{4-1}{1.3}+\frac{7-4}{4.7}+...+\frac{105-102}{102.105}\right)x=3\) \(\Leftrightarrow\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+..+\frac{1}{102}-\frac{1}{105}\right)x=3\)
\(\Leftrightarrow\left(1-\frac{1}{105}\right)x=3\Leftrightarrow\frac{104}{105}.x=3\Leftrightarrow x=\frac{315}{104}\)
\(3B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}.\)
\(3B=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{103-100}{100.103}\)
\(3B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}=1-\frac{1}{103}=\frac{102}{103}\)
\(B=\frac{102}{3.103}=\frac{34}{103}\)
a)\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{46}-\frac{1}{56}\)
=\(1-\frac{1}{56}=\frac{55}{56}\)
b)\(A.\frac{1}{3}=\frac{1}{3}.\left(\frac{3}{1.2}+\frac{3}{2.3}+....+\frac{3}{99.100}\right)\)
= \(\frac{1}{3}A=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{3}{99.100}\)
=> \(\frac{1}{3}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
=> \(\frac{1}{3}A=1-\frac{1}{100}=\frac{99}{100}\)
=> \(A=\frac{99}{100}.3=\frac{297}{100}\)
c)\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)
=\(1-\frac{1}{103}=\frac{102}{103}\)
d) \(\frac{3}{5}C=\frac{3}{5}.\left(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\right)\)
=\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\)
=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103}\)
=\(1-\frac{1}{103}=\frac{102}{103}\)
=>\(C=\frac{102}{103}.\frac{5}{3}=\frac{170}{103}\)
e) \(\frac{4}{7}D=\frac{4}{7}.\left(\frac{7}{1.5}+\frac{7}{5.9}+...+\frac{7}{101.105}\right)\)
=\(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{101.105}\)
=\(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)
=\(1-\frac{1}{105}=\frac{104}{105}\)
=< D=\(\frac{104}{105}.\frac{7}{4}=\frac{26}{15}\)
\(S_1=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{48\cdot49}+\frac{1}{49\cdot50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
\(S_2=\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+....+\frac{1}{94\cdot97}+\frac{1}{97\cdot100}\)
\(3S_2=\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+....+\frac{3}{94\cdot97}+\frac{3}{97\cdot100}\)
\(=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{97}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}=\frac{6}{25}\)
=> \(S_2=\frac{6}{25}:3=\frac{2}{25}\)
\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2015.2017}\)
\(A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
\(A=1-\dfrac{1}{2017}=\dfrac{2016}{2017}\)
\(B=\dfrac{3}{1.4}+\dfrac{3}{5.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)
\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(B=1-\dfrac{1}{103}=\dfrac{102}{103}\)
\(C=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{62.65}\)
\(3C=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{62.65}\right)\)
\(3C=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{62.65}\)
\(3C=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{62}-\dfrac{1}{65}\)
\(3C=\dfrac{1}{2}-\dfrac{1}{65}\)
\(3C=\dfrac{63}{130}\)
\(C=\dfrac{63}{130}:3=\dfrac{21}{130}\)
Ta có: \(B=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{3}-\frac{1}{8}\)
\(=\frac{5}{24}\)
Vậy \(B=\frac{5}{24}\)
Ta có: \(C=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{4}-\frac{1}{9}\)
\(=\frac{5}{36}\)
Vậy \(C=\frac{5}{36}\)
1)
A= \(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{39.40}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...+\frac{1}{39}-\frac{1}{40}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{40}\)
=> A= 27/120
A = \(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{39.40}\)
= \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{39}-\frac{1}{40}\)
= \(\frac{1}{3}-\frac{1}{40}\)
= \(\frac{37}{120}\)
B = \(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{37.40}\)
= \(\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{37}-\frac{1}{40}\right)\)
= \(\frac{1}{3}\left(\frac{1}{4}-\frac{1}{40}\right)\)
= \(\frac{1}{3}.\frac{9}{40}=\frac{3}{40}\)
C = \(\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{37.40}\)
= \(\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{37}-\frac{1}{40}\right)\)
= \(\frac{2}{3}.\left(\frac{1}{4}-\frac{1}{40}\right)\)
= \(\frac{2}{3}.\frac{9}{40}=\frac{3}{20}\)