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\(M=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(M=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)
\(M=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(M=1-\dfrac{1}{10^2}< 1\left(đpcm\right)\)
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,\(\dfrac{\left(0,9\right)^5}{\left(0,3\right)^6}=\dfrac{\left(0,3.3\right)^5}{\left(0,3\right)^6}=\dfrac{\left(0,3\right)^5.3^5}{\left(0,3\right)^6}=\dfrac{3^5}{\left(0,3\right)}\)
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= \(\dfrac{3}{5}\left(1-\dfrac{1}{105}\right)\)=\(\dfrac{3}{5}.\dfrac{104}{105}=\dfrac{312}{525}\)
Ta có:
\(A=\frac{3}{1\cdot5}+\frac{3}{5\cdot10}+...+\frac{3}{100\cdot105}\)
\(=\frac{3}{5}\cdot\left(\frac{5}{1\cdot5}+\frac{5}{5\cdot10}+...+\frac{5}{100\cdot105}\right)\)
\(=\frac{3}{5}\cdot\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{105}\right)\)
\(=\frac{3}{5}\left(1-\frac{1}{105}\right)=\frac{3}{5}\cdot\frac{104}{105}=\frac{312}{525}\)
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, \(\dfrac{1}{3}-2.1=-\dfrac{5}{3}\)
Câu b, B=\(\dfrac{5}{1\cdot2}+\dfrac{13}{2\cdot3}+\dfrac{25}{3\cdot4}+...+\dfrac{181}{9\cdot10}\)
\(=\left(\dfrac{1}{1\cdot2}+\dfrac{4}{1\cdot2}\right)+\left(\dfrac{1}{2\cdot3}+\dfrac{12}{2\cdot3}\right)+\left(\dfrac{1}{3\cdot4}+\dfrac{24}{3\cdot4}\right)+...+\left(\dfrac{1}{9\cdot10}+\dfrac{180}{9\cdot10}\right)\)=\(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)+\left(\dfrac{4}{1\cdot2}+\dfrac{12}{2\cdot3}+...+\dfrac{180}{9\cdot10}\right)\)
=\(\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)\)\(+\left(2+2+2+.......+2\right)\)
=\(\dfrac{1}{1}-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-......-\left(\dfrac{1}{9}-\dfrac{1}{9}\right)+\dfrac{1}{10}+\left(2\cdot9\right)\)
=\(1-\dfrac{1}{10}+18\) \(=\dfrac{9}{10}+18\)
=18.9
a, \(\dfrac{\dfrac{3}{2}-\dfrac{2}{5}+\dfrac{1}{10}}{\dfrac{3}{2}-\dfrac{2}{3}+\dfrac{1}{12}}=\dfrac{\dfrac{15}{10}-\dfrac{4}{10}+\dfrac{1}{10}}{\dfrac{18}{12}-\dfrac{8}{12}+\dfrac{1}{12}}=\dfrac{\dfrac{15-4+1}{10}}{\dfrac{18-8+1}{12}}=\dfrac{\dfrac{12}{10}}{\dfrac{11}{12}}=\dfrac{72}{55}\)
a) = (\(-\dfrac{141}{20}\)- \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{15}\) - \(\dfrac{1}{15}\)
= \(-\dfrac{73}{10}\) : - 5
= \(\dfrac{73}{50}\)
b) = \(\left(\dfrac{3}{25}-\dfrac{28}{25}\right)\). \(\dfrac{7}{3}\) : \(\left(\dfrac{7}{2}-\dfrac{11}{3}.14\right)\)
= \(-\dfrac{7}{3}\) . \(-\dfrac{6}{287}\)
= \(\dfrac{2}{41}\)
\(B=\dfrac{5}{1.2}+\dfrac{13}{2.3}+\dfrac{25}{3.4}+\dfrac{41}{4.5}+...+\dfrac{181}{9.10}\)
\(=\left(\dfrac{1}{1.2}+\dfrac{4}{1.2}\right)+\left(\dfrac{1}{2.3}+\dfrac{12}{2.3}\right)+\left(\dfrac{1}{3.4}+\dfrac{24}{3.4}\right)+...+\left(\dfrac{1}{9.10}+\dfrac{180}{9.10}\right)\)
\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\right)+\left(\dfrac{4}{1.2}+\dfrac{12}{2.3}+...+\dfrac{180}{9.10}\right)\)
\(=\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)+\left(2+2+...+2\right)\)
\(=1-\dfrac{1}{10}+\left(2.9\right)\)
\(=1-\dfrac{1}{10}+18\)
\(=\dfrac{9}{10}+18\)
\(=18\dfrac{9}{10}\)