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Hình như mik chưa tính nhưng vế sau là = 0 nên bnj ko cần tính vế trước đâu
( \(\frac{12}{199}\)+ \(\frac{23}{200}\)- \(\frac{34}{201}\)) x ( \(\frac{1}{2}\)- \(\frac{1}{3}\)- \(\frac{1}{6}\))
= ( \(\frac{12}{199}\)+ \(\frac{23}{200}\)- \(\frac{34}{201}\)) x ( \(\frac{3}{6}\)- \(\frac{2}{6}\)- \(\frac{1}{6}\))
=( \(\frac{12}{199}\) + \(\frac{23}{200}\) - \(\frac{34}{201}\)) x 0
= 0
Học tốt ^-^
vì\(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}=\frac{3}{6}-\frac{2}{6}-\frac{1}{6}=\frac{3-2-1}{6}=\frac{0}{6}=0\)
=> \(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot0=0\)
a) 7/13.7/15 - 5/12.21/39 + 49/91.8/15
= 7/13. 7/15 - 5/12. 7/13 + 7/13.8/15
= 7/13. ( 7/15 - 5/12 + 8/15)
= 7/13. ( 7/15 + 8/15 - 5/12)
= 7/13. ( 1 - 5/12)
= 7/13. 7/12
= 49/156
b) ( 12/199 + 23/100 - 34/201) . ( 1/2-1/3-1/6)
= ( 12/199 + 23/100 - 34/201).0
= 0
a) \(=\frac{7}{13}.\frac{7}{15}-\frac{5}{12}.\frac{7}{13}+\frac{7}{130}.\frac{8}{15}=\frac{7}{13}\left(\frac{7}{15}+\frac{8}{15}-\frac{5}{12}\right)=\frac{7}{13}\left(1-\frac{5}{12}\right)=\frac{7}{13}.\frac{7}{12}=\frac{48}{156}\)
b) \(=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).0=0\)
a)\(=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}\)
\(=\left(\frac{-3}{7}+\frac{3}{7}\right)-\left(\frac{15}{26}+\frac{2}{13}\right)\)
\(=0-\frac{19}{26}\)
\(=-\frac{19}{26}\)
c)\(=\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)
\(=\frac{-11}{23}.2-\frac{1}{23}\)
\(=\frac{-22}{23}-\frac{1}{23}\)
\(=-1\)
Ta có: \(B=\frac{1}{199}+\frac{2}{198}+...+\frac{199}{1}\)
\(=\frac{200-199}{199}+\frac{200-198}{198}+...+\frac{200-1}{1}\)
\(=\frac{200}{199}-\frac{199}{199}+\frac{200}{198}-\frac{198}{198}+...+\frac{200}{1}-\frac{1}{1}\)
\(=\left(\frac{200}{199}+\frac{200}{198}+...+\frac{200}{1}\right)-\left(\frac{199}{199}+\frac{198}{198}+...+\frac{1}{1}\right)\)
\(=200+200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)-199\)
\(=200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)+\frac{200}{200}\)
\(=200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)}=\frac{1}{200}\)
Ta có :
\(B=\frac{1}{199}+\frac{2}{198}+....+\frac{198}{2}+\frac{199}{1}\)
\(B=1+\frac{1}{199}+1+\frac{1}{198}+....+1+\frac{198}{2}\)
\(B=\frac{200}{199}+\frac{200}{198}+...+\frac{200}{2}\)
\(B=200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)}=\frac{1}{200}\)
Vậy \(\frac{A}{B}=\frac{1}{200}\)
a, 3 \(\frac{14}{19}\)+ \(\frac{13}{17}\)+ \(\frac{35}{43}\)+ 6\(\frac{5}{19}\)+ \(\frac{8}{43}\)= \(\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}=\)\(9+1+\frac{13}{17}=8+\frac{13}{17}=8\frac{13}{17}\)
b, \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1\frac{5}{7}\)\(=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}\right)+1\frac{5}{7}\)\(=\frac{-5}{7}.1+1\frac{5}{7}\)\(=\frac{-5}{7}+\frac{12}{7}=\frac{7}{7}=1\)
Chúc bn học tốt
\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\frac{5}{19}+\frac{8}{43}\)
\(=\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}\)
\(=10+1+\frac{13}{17}=11+\frac{13}{17}=11\frac{13}{17}\)
a)\(=\frac{2}{3}\left(\frac{2013}{2012}-\frac{1}{2012}\right)+\frac{1}{3}\)
\(=\frac{2}{3}\times1+\frac{1}{3}\)
\(=\frac{3}{3}=1\)
b)\(=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).\left(\frac{3-2-1}{6}\right)\)
\(=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).0\)
=0