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Trả lời
b)(1/3+12/67+13/41)-(79/67-28/41)
=1/3+12/67+13/41-79/67+28/41
=1/3+(12/67-79/67)+(13/41+28/41)
=1/3+(-67/67)+41/41
=1/3+(-1)+1
=1/3+0
=1/3.
a, \(\left(31\frac{6}{13}+5\frac{9}{41}\right)-36\frac{6}{13}=31\frac{6}{16}+5\frac{9}{41}-36\frac{6}{13}\)
\(=\left(31\frac{6}{16}-31\frac{6}{16}\right)+5\frac{9}{41}\)
\(=0+5\frac{9}{41}=5\frac{9}{41}\)
b, \(\left(17\frac{29}{31}-3\frac{7}{8}\right)-\left(2\frac{28}{31}-4\right)=17\frac{9}{31}-3\frac{7}{8}-2\frac{28}{31}+4\)
\(a,\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)< x< \left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}\)
\(taco:\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)=\frac{35}{36}\cdot\frac{-36}{35}=-1\)
\(\left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}=\frac{13}{8}\cdot\frac{8}{13}=1\)
\(=>x=0\)
\(b,\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}< x< \frac{-1}{2}+2+\frac{5}{2}\)(dau <co dau gach ngang o duoi nha)
\(taco:\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}=\frac{-5}{6}+\frac{8}{3}+\frac{-29}{3}=\frac{-5}{6}+\frac{16}{6}+\frac{-58}{6}=\frac{-47}{6}=-7,8\)
\(\frac{-1}{2}+2+\frac{5}{2}=\frac{3}{2}+\frac{5}{2}=4\)
tu do \(=>x=-7,8;...;0;1;2;3;4\)
Câu a:
(4 5/37 - 3 4/5 + 8 15/29) - (3 5/37 - 6 14/29)
= (4 5/37 - 3 5/37) + (8 15/29 + 6 14/29) - 3 4/5
= 1 + 15 - 3 4/5
= 16 - 3 - 4/5
= 13 - 4/5
= 61/5
Câu b:
(2\(\frac56\) + 1\(\frac49\)) : (10\(\frac{1}{12}\) - 9\(\frac12\))
= (17/6 + 13/9) :(121/12 - 19/2)
= (51/18 + 26/18) :(121/12 - 114/2)
= 77/18 : 7/12
= 77/18 x 12/7
= 22/3
\(1\frac57-\frac57\times\frac{2}{11}-\frac57\times\frac{9}{11}\)
=\(\frac57\times\left(1-\frac{2}{11}-\frac{9}{11}\right)\)
=\(\frac57\times\left(\frac{11}{11}-\frac{2}{11}-\frac{9}{11}\right)\)
=\(\frac57\times1\)
=\(\frac57\)
\(\frac{5}{11}\times\frac{18}{29}-\frac{5}{11}\times\frac{8}{29}+\frac{5}{11}\times\frac{19}{26}\)
=\(\frac{5}{11}\times\left(\frac{18}{29}-\frac{8}{29}+\frac{19}{29}\right)\)
=\(\frac{5}{11}\times1\)
=\(\frac{5}{11}\)
\(\frac{\frac{3}{67}.\left(17\frac{21}{56}-13\frac{21}{45}\right):\left(\frac{3}{5.22}+\frac{54}{44.65}+\frac{18}{65.72}\right)}{29^3:100-29^3:0,47}\)
\(=\frac{\frac{3}{67}\left(\frac{139}{8}-\frac{202}{15}\right):\left(\frac{3}{110}+\frac{54}{2860}+\frac{18}{4680}\right)}{29^3.\frac{1}{100}-29^3.\frac{47}{100}}\)
\(=\frac{\frac{3}{67}.\frac{469}{120}:\frac{1}{20}}{29^3\left(\frac{1}{100}-\frac{47}{100}\right)}\)
\(=\frac{\frac{7}{40}.20}{29^3.\left(-\frac{23}{50}\right)}\).
\(=\frac{\frac{7}{2}}{-11218,94}\)
\(=-\frac{175}{560947}\)