\(\dfrac{1}{13}+\dfrac{3}{13\cdot23}+\dfrac{3}{23\cdot33}+\dfrac{3}{33\cdot43}+...+\dfrac{3}{1993\cdot2003}\)
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a)-91 phần200
b)-25phần 4
c)5 phần 2
d)2
e)0
a, ( 0,36-2,18) : ( 3,8 + 0,2)
= -1,82 : 4
=-0,455 hay -91/200
b, 3/8*19/1/3-3/8*33/1/3
=3/8*(19/1/3-33/1/3)
=3/8*(-14)
=-21/4
\(A=\dfrac{115-\dfrac{10}{7}-\dfrac{5}{11}+\dfrac{5}{23}}{403-\dfrac{26}{7}-\dfrac{13}{11}+\dfrac{13}{23}}+\dfrac{\dfrac{3}{5}+\dfrac{3}{13}-0,9}{\dfrac{7}{91}+0,2-\dfrac{3}{30}}\)
\(=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\)
\(=\dfrac{5}{13}+3\)
\(=\dfrac{5}{13}+\dfrac{39}{13}\)
\(=\dfrac{44}{13}\)
\(=3\dfrac{5}{13}\)
CHÚC BN HC TỐT
\(A=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\)
=5/13+3
=39/13+5/13=44/13
a: =27/45-20/45=7/45
b: \(=\dfrac{3}{5}+\dfrac{30}{40}=\dfrac{3}{5}+\dfrac{3}{4}=\dfrac{12}{20}+\dfrac{15}{20}=\dfrac{27}{20}\)
c: \(=\dfrac{8}{13}\left(\dfrac{7}{2}-\dfrac{5}{2}+1\right)=\dfrac{8}{13}\cdot2=\dfrac{16}{13}\)
d: \(=\dfrac{9}{23}\left(\dfrac{5}{17}-\dfrac{22}{17}\right)+11+\dfrac{9}{23}=11\)
a) \(\dfrac{3}{5}+\dfrac{-4}{9}=\dfrac{27}{45}+\dfrac{-20}{45}=\dfrac{7}{45}\)
b) \(\dfrac{3}{5}+\dfrac{2}{5}.\dfrac{15}{8}=1.\dfrac{15}{8}=\dfrac{15}{8}\)
c) \(\dfrac{7}{2}.\dfrac{8}{13}+\dfrac{8}{13}.\dfrac{-5}{2}+\dfrac{8}{13}=\dfrac{8}{13}.\left(\dfrac{7}{2}+\dfrac{-5}{2}\right)=\dfrac{8}{13}.1=\dfrac{8}{13}\)
d) \(\dfrac{-5}{17}.\dfrac{-9}{23}+\dfrac{9}{23}.\dfrac{-22}{17}+11\dfrac{9}{23}=\dfrac{9}{23}.\left(\dfrac{-5}{17}+\dfrac{-22}{17}\right)=\dfrac{-243}{391}\)
\(\dfrac{24\cdot47-23}{24+47\cdot23}\cdot\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{\dfrac{9}{1001}-\dfrac{9}{13}+\dfrac{9}{7}-\dfrac{9}{11}+9}\\ =\dfrac{24\cdot\left(24+23\right)-23}{24+\left(24+23\right)\cdot23}\cdot\dfrac{3\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}{9\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}\\ =\dfrac{24\cdot24+24\cdot23-23\cdot1}{24+24\cdot23+23\cdot23}\cdot\dfrac{1}{3}\\ =\dfrac{23\left(24-1\right)+24\cdot24}{24\left(1+23\right)+23\cdot23}\cdot\dfrac{1}{3}=\dfrac{23\cdot23+24\cdot24}{24\cdot24+23\cdot23}\cdot\dfrac{1}{3}\\ =1\cdot\dfrac{1}{3}=\dfrac{1}{3}\)
a) Ta có: \(\dfrac{-3}{7}+\dfrac{15}{26}-\left(\dfrac{2}{13}-\dfrac{3}{7}\right)\)
\(=\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}\)
\(=\dfrac{15}{26}-\dfrac{4}{26}\)
\(=\dfrac{11}{26}\)
b) Ta có: \(2\cdot\dfrac{3}{7}+\left(\dfrac{2}{9}-1\dfrac{3}{7}\right)-\dfrac{5}{3}:\dfrac{1}{9}\)
\(=\dfrac{6}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9\)
\(=\dfrac{-4}{7}+\dfrac{2}{9}-15\)
\(=\dfrac{-36}{63}+\dfrac{14}{63}-\dfrac{945}{63}\)
\(=\dfrac{-967}{63}\)
c) Ta có: \(\dfrac{-11}{23}\cdot\dfrac{6}{7}+\dfrac{8}{7}\cdot\dfrac{-11}{23}-\dfrac{1}{23}\)
\(=\dfrac{-11}{23}\cdot\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}\)
\(=\dfrac{-11}{23}\cdot2-\dfrac{1}{23}\)
\(=-1\)
d) Ta có: \(\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{24}\right)\)
\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{4}{24}-\dfrac{3}{24}-\dfrac{1}{24}\right)\)
\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot0\)
=0
Công thức tổng quát:n thuộc N
3/((10n+3)(10n+13))
=(3/10)((1/(10n+3))-(1/(10n+13)))
1/13 + 3/(13×23) + 3/(23.33) + ....+3/(1993.2003)
=(3/10)(1/3-1/13+1/13-1/23+...+1/1993-1/2003)
=(3/10)(1/3-1/2003)
=200/2003