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2 ) B = \(1\frac{6}{41}.\left(\frac{12+\frac{12}{19}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2006}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2006}}\right).\frac{124242423}{237373735}\)
B = 47/41 . ( 12/3 : 4/5 ) . 123/235
B = 47/41 . ( 4 : 4/5 ) . 123/235
B = 47/41 . 5 . 123/235
B = \(\frac{47.5.123}{41.235}\)
B = 3
1 ) A = \(\frac{636363.37-373737.63}{1+2+3+...+2006}\)
A = \(\frac{63.10101.37-37.10101.63}{1+2+3+...+2006}\)
A = \(\frac{0}{1+2+3+...+2006}\)
A = 0
kazuto kirigaya thật là bt làm ko đó ko bt thì nói đi còn bt thì làm đi
A = -4/5x(1/2+1/3+1/4)= -4/5x1 = -4/5
B = 6/19 x ( 3/4+4/3+-1/2)= 6/19x 19 = 6
C = 2002/2003x(3/4+5/6-19/12)=2003/2002x0=0
P/s : nhìn thì khủng thật ! :v
\(B=81.\left[\frac{\left[12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}\right]}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right].\frac{158158158}{711711711}\)
\(B=81.\left[\frac{12.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right].\frac{158}{711}\)
\(B=81.\left(\frac{3}{1}:\frac{5}{6}\right).\frac{158}{711}\)
\(B=81.\frac{18}{5}.\frac{158}{711}\)
\(B=\frac{1458}{5}.\frac{158}{711}=\frac{324}{5}\)
Vậy \(B=\frac{324}{5}\)
a) Ta có: \(\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)
\(=\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{9}{4}\cdot\frac{8}{3}\)
\(=4\cdot\frac{-1}{3}\cdot\frac{4}{7}\cdot3\)
\(=12\cdot\frac{-4}{21}=\frac{-48}{21}=\frac{-16}{7}\)
b) Ta có: \(5\cdot\frac{7}{5}=\frac{35}{5}=7\)
c) Ta có: \(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}\)
\(=\frac{5}{9}\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)\)
\(=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)
d) Ta có: \(4\cdot11\cdot\frac{3}{4}\cdot\frac{9}{121}\)
\(=\frac{4\cdot11\cdot3\cdot9}{4\cdot121}=\frac{27}{11}\)
e) Ta có: \(\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}:\frac{-21}{20}\)
\(=\frac{4}{3}+\frac{4}{3}=\frac{8}{3}\)
g) Ta có: \(2\frac{1}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\left(\frac{2}{3}+0,4\cdot5\right)\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\frac{2}{3}+2\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\frac{7}{6}\)
\(=\frac{7}{3}-\frac{7}{18}=\frac{42}{18}-\frac{7}{18}=\frac{35}{18}\)
a) A = \(\frac{19}{23}.\frac{-4}{27}-\frac{4}{23}.\frac{2}{7}\)
= \(\frac{19}{7}.\frac{-4}{23}+\frac{-4}{23}.\frac{2}{7}\)
= \(\frac{-4}{23}.\left(\frac{19}{7}+\frac{2}{7}\right)\)
= \(\frac{-4}{23}.3\)
= \(\frac{-12}{23}\)
b) B = \(\frac{3}{5}+\frac{2}{5}.\frac{-11}{3}+\frac{2}{3}.\frac{-2}{5}+\frac{14}{15}\)
= \(\frac{9+14}{15}+\frac{2}{5}.\frac{-11}{3}+\frac{-2}{3}.\frac{2}{5}\)
= \(\frac{23}{15}+\frac{2}{5}\left(\frac{-11}{3}+\frac{-2}{3}\right)\)
= \(\frac{23}{15}+\frac{2}{5}.\frac{-13}{3}\)
= \(\frac{23}{15}+\frac{-26}{15}\)
= \(\frac{-3}{15}=\frac{-1}{5}\)
\(a,A=\frac{3}{5}.\frac{5}{4}-\frac{3}{5}.\frac{1}{4}\)
\(=\frac{3}{5}.\left(\frac{5}{4}-\frac{1}{4}\right)\)
\(=\frac{3}{5}.1=\frac{3}{5}\)
\(b,B=6,3+\left(-6,3\right)+4,9\)
\(=0+4,9=4,9\)