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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
a: =11+3/4-6-5/6+4+1/2+1+2/3
=10+9/12-10/12+6/12+8/12
=10+13/12=133/12
b: \(=2+\dfrac{17}{20}-1-\dfrac{11}{15}+2+\dfrac{3}{20}\)
=3-11/15
=34/15
c: \(=\dfrac{31}{7}:\left(\dfrac{7}{5}\cdot\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}=\dfrac{5}{7}\)
d: \(=\dfrac{29}{8}\cdot\dfrac{36}{29}\cdot\dfrac{15}{23}\cdot\dfrac{23}{5}=\dfrac{9}{2}\cdot3=\dfrac{27}{2}\)
a) 2011 + 5[300 - (17 - 7)2 ]
= 2011 + 5[300 - 100]
= 2011 + 5.200
= 2011 + 1000
= 3011
b) 695 - [200 + (11 - 1)2 ]
= 695 - [200 + 100]
= 695 - 300
= 395
c) 129 - 5[29 - (6-1)2 ]
= 129 - 5[29 - 25]
= 129 - 5 . 4
= 129 - 20
= 109
d) 2345 - 1000 : [19 - 2(21-18)2 ]
= 2345 - 1000 : [19 - 2 . 9]
= 2345 - 1000 : 1
= 2345 - 1000
= 1345
a, 2011+5 [300-(17-7)2 ] =2011+5[300-102]
=2011+5[300-100]=2011+5.200
=2011+1000=3011
b, 695-[200+(11-1)2]=695-[200+102]
=695-[200+100]=695-300=395