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a)\(\dfrac{2}{3}-\dfrac{3}{5}:\left(-1\dfrac{1}{5}\right)+\left(\dfrac{-2}{3}\right)\cdot\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{3}{5}\cdot\dfrac{-5}{6}+\left(\dfrac{-1}{4}\right)=\dfrac{5}{12}+\dfrac{1}{2}=\dfrac{11}{12}\)
b)\(17\dfrac{11}{9}-\left(6\dfrac{3}{13}+7\dfrac{11}{19}\right)+\left(10\dfrac{3}{13}-5\dfrac{1}{4}\right)=\dfrac{164}{9}-\left(\dfrac{81}{13}+\dfrac{144}{19}\right)+\left(\dfrac{133}{13}-\dfrac{21}{4}\right)=\dfrac{164}{9}-\dfrac{3411}{247}+\dfrac{259}{52}=\dfrac{6425}{684}\)
c)\(\left(\dfrac{-3}{2}\right)^2-\left[-2\dfrac{1}{3}-\left(\dfrac{3}{4}+\dfrac{1}{3}\right):2\dfrac{3}{5}\right]\cdot\left(\dfrac{-3}{4}\right)=\dfrac{9}{4}-\left[\dfrac{-7}{3}-\dfrac{13}{12}\cdot\dfrac{5}{13}\right]\cdot\left(\dfrac{-3}{4}\right)=\dfrac{9}{4}-\left(\dfrac{-11}{4}\right)\cdot\left(\dfrac{-3}{4}\right)=\dfrac{3}{16}\)
d)\(\dfrac{21}{33}:\dfrac{11}{5}-\dfrac{13}{33}:\dfrac{11}{5}+\dfrac{25}{33}:\dfrac{11}{5}+\dfrac{6}{11}=\dfrac{5}{11}\cdot\left(\dfrac{21}{33}-\dfrac{13}{33}+\dfrac{25}{33}\right)+\dfrac{6}{11}=\dfrac{5}{11}\cdot1+\dfrac{6}{11}=1\)
\(a)\dfrac{2}{3}-\dfrac{3}{5}:\left(-1\dfrac{1}{5}\right)+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{3}{5}:\left(\dfrac{-6}{5}\right)+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{-1}{2}+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{-1}{2}+\dfrac{-1}{4}\)
\(=\dfrac{7}{6}+\dfrac{-1}{4}\)
\(=\dfrac{11}{12}\)
\(\left(1+\dfrac{7}{9}\right).\left(1+\dfrac{7}{20}\right).\left(1+\dfrac{7}{33}.\right)\left(1+\dfrac{7}{48}\right)...\left(1+\dfrac{7}{180}\right)\)
\(=\dfrac{16}{9}.\dfrac{27}{20}.\dfrac{40}{33}.\dfrac{55}{48}...\dfrac{7}{180}\)
\(=\dfrac{2.8}{1.9}.\dfrac{3.9}{2.10}.\dfrac{4.10}{3.11}.\dfrac{5.11}{4.12}...\dfrac{11.17}{10.18}\)
\(=\dfrac{\left(2.3.4.5...11\right).\left(8.9.10.11...17\right)}{\left(1.2.3.4...10\right).\left(9.10.11.12...18\right)}\)
\(=\dfrac{11.8}{1.18}=\dfrac{88}{18}=\dfrac{44}{9}\)
ta có ;
\(\left(1+\dfrac{7}{9}\right)\cdot\left(1+\dfrac{7}{20}\right).\left(1+\dfrac{7}{33}\right)...\left(1+\dfrac{1}{180}\right)\)
=\(\dfrac{16}{9}.\dfrac{27}{20}.\dfrac{40}{33}....\dfrac{187}{180}\)
=\(\dfrac{8.2}{9.1}.\dfrac{9.3}{10.2}.\dfrac{10.4}{3.11}.\dfrac{11.5}{4.12}....\dfrac{17.11}{18.10}\)
=\(\dfrac{8.9.10.11.12.13.14.15.16.17.2.3.4.5.6.7.8.9.10.11}{9.10.11.12.13.14.15.16.17.18.1.2.3.4.5.6.7.8.9.10}\)
=\(\dfrac{8.11}{18}=\dfrac{88}{18}=\dfrac{44}{9}\)
\(A=\dfrac{-19}{9}.\dfrac{1}{2}-\dfrac{4}{11}.\dfrac{-11}{9}+\left(-\dfrac{2}{3}\right)=-\dfrac{23}{18}\)
\(B=\left(-\dfrac{15}{6}\right):\dfrac{-1}{2}+\dfrac{7}{-12}-\dfrac{1}{3}.\dfrac{-11}{2}=\dfrac{25}{4}\)
\(C=\dfrac{3}{4}.\left(-8\right)-\dfrac{1}{3}.\dfrac{-7}{2}-\dfrac{5}{18}=-\dfrac{46}{9}\)
\(A=\dfrac{-19}{18}+\dfrac{4}{9}-\dfrac{2}{3}=\dfrac{-19}{18}+\dfrac{8}{18}-\dfrac{12}{18}=\dfrac{-23}{18}\)
\(B=\dfrac{-5}{2}\cdot\dfrac{-2}{1}-\dfrac{7}{12}+\dfrac{11}{6}=\dfrac{5\cdot12-7+22}{12}=\dfrac{75}{12}=\dfrac{25}{4}\)
A = - 522 - { - 222 - [ - 122 - (100 - 522) + 2022] }
A = - 522 - { -222 - [- 122 - 100 + 522 ] + 2022}
A = - 522 - { -222 - { - 222 + 522 } + 2022}
A = - 522 - {- 222 + 222 - 522 + 2022}
A = -522 + 522 - 2022
A = - 2022
B = 1 + \(\dfrac{1}{2}\)(1 + 2) + \(\dfrac{1}{3}\).(1 + 2 + 3) + ... + \(\dfrac{1}{20}\).(1 + 2+ 3 + ... + 20)
B = 1+\(\dfrac{1}{2}\)\(\times\)(1+2)\(\times\)[(2-1):1+1]:2+ ... + \(\dfrac{1}{20}\)\(\times\) (20 + 1)\(\times\)[(20-1):1+1]:2
B = 1 + \(\dfrac{1}{2}\) \(\times\) 3 \(\times\) 2:2 + \(\dfrac{1}{3}\) \(\times\)4 \(\times\) 3 : 2+....+ \(\dfrac{1}{20}\) \(\times\)21 \(\times\) 20 : 2
B = 1 + \(\dfrac{3}{2}\) + \(\dfrac{4}{2}\) + ....+ \(\dfrac{21}{2}\)
B = \(\dfrac{2+3+4+...+21}{2}\)
B = \(\dfrac{\left(21+2\right)\left[\left(21-2\right):1+1\right]:2}{2}\)
B = \(\dfrac{23\times20:2}{2}\)
B = \(\dfrac{23\times10}{2}\)
B = 23
a) Ta có: \(\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+\dfrac{20}{-17}+\dfrac{-2}{9}+\dfrac{21}{56}\)
\(=\left(\dfrac{3}{17}-\dfrac{20}{17}\right)+\left(\dfrac{2}{9}-\dfrac{2}{9}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)
\(=-1+1=0\)
b) Ta có: \(\left(\dfrac{9}{16}+\dfrac{8}{-27}\right)+\left(1+\dfrac{7}{16}+\dfrac{-19}{27}\right)\)
\(=\left(\dfrac{9}{16}+\dfrac{7}{16}\right)+\left(\dfrac{-8}{27}-\dfrac{19}{27}\right)+1\)
=1-1+1=1
ta có : \(\dfrac{3}{2}\)A= \(\dfrac{3}{4}+\)\(\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\)\(...+\left(\dfrac{3}{2}\right)^{2013}\) (1)
A= \(\dfrac{1}{2}+\dfrac{3}{2}\)\(+\left(\dfrac{3}{2}\right)^2+...+\)\(\left(\dfrac{3}{2}\right)^{2012}\) (2)
Lấy (1) trừ đi (2) vế theo vế:
\(\dfrac{3}{2}A-A=\dfrac{3}{4}-\dfrac{1}{2}-\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^{2013}\)
\(\dfrac{1}{2}A=\left(\dfrac{3}{2}\right)^{2013}-\dfrac{5}{4}\Rightarrow A=\dfrac{3^{2013}}{2^{2012}}-\dfrac{5}{2}\)
ta có : \(B=\left(\dfrac{3}{2}\right)^{2013}:2=\dfrac{3^{2013}}{2^{2013}}.\dfrac{1}{2}=\dfrac{3^{2013}}{2^{2014}}\)
Vậy \(A-B=\dfrac{3^{2013}}{2^{2014}}-\left(\dfrac{3^{2013}}{2^{2012}}-\dfrac{5}{2}\right)\)
13: \(=\dfrac{4}{9}\cdot\left(-7\right)+\left(6+\dfrac{5}{9}\right)\cdot\left(-7\right)\)
\(=\left(-7\right)\left(\dfrac{4}{9}+6+\dfrac{5}{9}\right)=\left(-7\right)\cdot7=-49\)
14: \(=\left(\dfrac{-3}{4}+\dfrac{5}{13}\right)\cdot\dfrac{7}{2}-\left(\dfrac{9}{4}+\dfrac{8}{13}\right)\cdot\dfrac{7}{2}\)
\(=\dfrac{7}{2}\left(-\dfrac{3}{4}+\dfrac{5}{13}-\dfrac{9}{4}-\dfrac{8}{13}\right)\)
\(=\dfrac{7}{2}\left(-3-\dfrac{3}{13}\right)=\dfrac{7}{2}\cdot\dfrac{-42}{13}=\dfrac{-147}{13}\)
a: \(=\left(\dfrac{17}{10}+\dfrac{70}{10}-\dfrac{87}{10}\right):\left(\dfrac{23}{4}-\dfrac{11}{4}+\dfrac{9}{25}\right)\cdot\left(12,98\cdot0,25\right)+12,5\)
\(=0:\left(3+\dfrac{9}{25}\right)\cdot\left(12,98+0,25\right)+12,5\)
=12,5
b: \(=\dfrac{13}{12}\cdot\dfrac{27}{5}\cdot2\cdot\dfrac{34}{9}\cdot2\cdot\dfrac{2}{17}\)
\(=\dfrac{13}{12}\cdot2\cdot\dfrac{27}{5}\cdot\dfrac{34}{9}\cdot\dfrac{4}{17}\)
\(=\dfrac{13}{6}\cdot\dfrac{27}{5}\cdot\dfrac{8}{9}=\dfrac{8}{6}\cdot3\cdot\dfrac{13}{5}=4\cdot\dfrac{13}{5}=\dfrac{52}{5}\)
9/48.(-2,4)+9/10:2
=-9/20+9/10:2
=-9/20+9/20
=0