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Ta có:
\(A=\frac{3}{1\cdot5}+\frac{3}{5\cdot10}+...+\frac{3}{100\cdot105}\)
\(=\frac{3}{5}\cdot\left(\frac{5}{1\cdot5}+\frac{5}{5\cdot10}+...+\frac{5}{100\cdot105}\right)\)
\(=\frac{3}{5}\cdot\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{105}\right)\)
\(=\frac{3}{5}\left(1-\frac{1}{105}\right)=\frac{3}{5}\cdot\frac{104}{105}=\frac{312}{525}\)
\(B=\dfrac{5}{1.2}+\dfrac{13}{2.3}+\dfrac{25}{3.4}+\dfrac{41}{4.5}+...+\dfrac{181}{9.10}\)
\(=\left(\dfrac{1}{1.2}+\dfrac{4}{1.2}\right)+\left(\dfrac{1}{2.3}+\dfrac{12}{2.3}\right)+\left(\dfrac{1}{3.4}+\dfrac{24}{3.4}\right)+...+\left(\dfrac{1}{9.10}+\dfrac{180}{9.10}\right)\)
\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\right)+\left(\dfrac{4}{1.2}+\dfrac{12}{2.3}+...+\dfrac{180}{9.10}\right)\)
\(=\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)+\left(2+2+...+2\right)\)
\(=1-\dfrac{1}{10}+\left(2.9\right)\)
\(=1-\dfrac{1}{10}+18\)
\(=\dfrac{9}{10}+18\)
\(=18\dfrac{9}{10}\)
S = 1.3 + 3.5 + 5.7 + ...+ 99.101
=>6S = 1.3.6 + 3.5.6 + 5.7.6 + ...+ 99.101.6
6S = 1.3.(5+1) + 3.5.(7-1) + 5.7.(9-3) + ...+ 99.101.(103-97)
6S = 1.3.5 + 1.3 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + ...+ 99.101.103 - 97.99.101
6S = 1.3 + 99.101.103
S = 171 650
S = 1.3 + 3.5 + 5.7 + ...+ 99.101
=>6S = 1.3.6 + 3.5.6 + 5.7.6 + ...+ 99.101.6
6S = 1.3.(5+1) + 3.5.(7-1) + 5.7.(9-3) + ...+ 99.101.(103-97)
6S = 1.3.5 + 1.3 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + ...+ 99.101.103 - 97.99.101
6S = 1.3 + 99.101.103
S = 171 650
a, \(\dfrac{20^5.5^{10}}{100^5}=\dfrac{20^5.5^{10}}{\left(20.5\right)^5}=\dfrac{20^5.5^{10}}{20^5.5^5}=5^5\)
b,\(\dfrac{\left(0,9\right)^5}{\left(0,3\right)^6}=\dfrac{\left(0,3.3\right)^5}{\left(0,3\right)^6}=\dfrac{\left(0,3\right)^5.3^5}{\left(0,3\right)^6}=\dfrac{3^5}{\left(0,3\right)}\)
\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+....+\dfrac{3}{32\cdot35}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+....+\dfrac{1}{32}-\dfrac{1}{35}\)
\(=\dfrac{1}{2}-\dfrac{1}{35}\)
\(=\dfrac{33}{70}\)
a,
\(\dfrac{\left(3^3\right)^{15}.5^3.\left(2^3\right)^4}{\left(5^2\right)^2.\left(3^4\right)^{11}.2^{11}}=\dfrac{3^{45}.5^3.2^{12}}{5^4.3^{44}.2^{11}}=\dfrac{6}{5}\)
b, \(\left(-\dfrac{14}{25}\right)^2.\dfrac{125}{49}+\left(-3\dfrac{11}{36}\right).2\dfrac{2}{17}=\dfrac{4}{5}.\left(-7\right)=-\dfrac{28}{5}\)
c, \(\dfrac{1}{3}-2.1=-\dfrac{5}{3}\)
(31/5:217/10-1,2*125)*7/4-1/2
=(2/7-150)*7/4-1/2
=-1048/7*7/4-1/2
=-262-1/2
=-525/2
\(\left(6\dfrac{1}{5}:21,7-1,2\cdot5^3\right)\cdot1\dfrac{3}{4}-\dfrac{1}{2}\)
\(=\left(\dfrac{31}{5}\cdot\dfrac{10}{217}-1,2\cdot125\right)\cdot1,75-0,5\)
\(=\left(\dfrac{62}{217}-150\right)\cdot1,75-0,5\)
\(=\dfrac{62}{217}\cdot1,75-150\cdot1,75-0,5\)
\(=0,5-262,5-0,5\)
\(=-262,5\)
Ta có: B-A=1x3+2x4+3x5+4x6+...+100x102-(1x2+2x3+3x4+4x5+...+100x101)
=1x3+2x4+3x5+4x6+...100x102-1x2-2x3-3x4-4x5-...-100x101
=1+2+3+4+...+100
=((100-1):1+1)x((100-1):2)
=100x(101:2)
=5050
Có: A=\(\dfrac{3}{1.5}+\dfrac{3}{5.10}+...+\dfrac{3}{100.105}\)
=> A=\(3.\dfrac{5}{5}\left(\dfrac{1}{1.5}+\dfrac{1}{5.10}+...+\dfrac{1}{100.105}\right)\)
=> A= \(3.\dfrac{1}{5}\left(\dfrac{5}{1.5}+\dfrac{5}{5.10}+...+\dfrac{5}{100.105}\right)\)
=> A=\(\dfrac{3}{5}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{105}\right)\)
=> A= \(\dfrac{3}{5}\left(1-\dfrac{1}{105}\right)\)=\(\dfrac{3}{5}.\dfrac{104}{105}=\dfrac{312}{525}\)