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\(\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{5}\right)\times\left(1-\frac{1}{6}\right)\times\left(1-\frac{1}{7}\right)\times\left(1-\frac{1}{8}\right)-\frac{1}{4}\times\frac{1}{2}\)
\(=\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}\times\frac{6}{7}\times\frac{7}{8}-\frac{1}{4}\times\frac{1}{2}\)
\(=\frac{2}{8}-\frac{1}{4}\times\frac{1}{2}\)
\(=\frac{2}{8}-\frac{1}{8}=\frac{1}{8}\)
\(a.\frac{3}{7}.\frac{5}{6}+\frac{5}{6}.\frac{4}{7}=\left(\frac{3}{7}+\frac{4}{7}\right).\frac{5}{6}=1.\frac{5}{6}=\frac{5}{6}\)
\(b.\frac{1}{6}:\frac{4}{5}+\frac{1}{8}:\frac{4}{5}=\left(\frac{1}{6}+\frac{1}{8}\right):\frac{4}{5}=\left(\frac{4}{24}+\frac{3}{24}\right):\frac{4}{5}=\frac{7}{24}:\frac{4}{5}=\frac{7.5}{24.4}\)\(=\frac{35}{96}\)
\(c.\frac{1}{6}:\frac{4}{5}-\frac{1}{8}:\frac{4}{5}=\left(\frac{1}{6}-\frac{1}{8}\right):\frac{4}{5}=\left(\frac{4}{24}-\frac{3}{24}\right):\frac{4}{5}\)\(=\frac{1}{24}:\frac{4}{5}=\frac{1.5}{24.4}=\frac{5}{96}\)
a) \(\frac{3}{7}\cdot\frac{5}{6}+\frac{5}{6}\cdot\frac{4}{7}\)
\(=\left(\frac{3}{7}+\frac{4}{7}\right)\cdot\frac{5}{6}\)
\(=1\cdot\frac{5}{6}\)
\(=\frac{5}{6}\)
b) \(\frac{1}{6}:\frac{4}{5}+\frac{1}{8}:\frac{4}{5}\)
\(=\left(\frac{1}{6}+\frac{1}{8}\right):\frac{4}{5}\)
\(=\frac{7}{24}:\frac{4}{5}\)
\(=\frac{35}{96}\)
c) \(\frac{1}{6}:\frac{4}{5}-\frac{1}{8}:\frac{4}{5}\)
\(=\left(\frac{1}{6}-\frac{1}{8}\right):\frac{4}{5}\)
\(=\frac{1}{24}:\frac{4}{5}\)
\(=\frac{5}{96}\)
a) \(\frac{3}{5}\times y+\frac{1}{2}:\frac{5}{3}-\frac{5}{4}=\frac{1}{2}\times\frac{1}{3}\)
\(\Rightarrow\frac{3}{5}\times y+\frac{3}{10}-\frac{5}{4}=\frac{1}{6}\)
\(\Rightarrow\frac{3}{5}\times y+\left(-\frac{19}{20}\right)=\frac{1}{6}\)
\(\Rightarrow\frac{3}{5}\times y=\frac{67}{60}\)
\(\Rightarrow y=\frac{67}{36}\)
b) \(\frac{4}{5}:y+\frac{1}{4}\times\frac{1}{6}-\frac{1}{2}=\frac{1}{3}\times\frac{5}{2}\)
\(\Rightarrow\frac{4}{5}:y+\frac{1}{24}-\frac{1}{2}=\frac{5}{6}\)
\(\Rightarrow\frac{4}{5}:y+\left(-\frac{11}{24}\right)=\frac{5}{6}\)
\(\Rightarrow\frac{4}{5}:y=\frac{5}{6}+\frac{11}{24}=\frac{31}{24}\)
\(\Rightarrow y=\frac{4}{5}:\frac{31}{24}=\frac{96}{155}\)
c) \(\frac{3}{5}\times y-\frac{4}{5}:3+\frac{1}{12}=\frac{3}{2}+\frac{1}{5}\)
\(\Rightarrow\frac{3}{5}\times y-\frac{4}{15}+\frac{1}{12}=\frac{17}{10}\)
\(\Rightarrow\frac{3}{5}\times y-\frac{4}{15}=\frac{97}{60}\)
\(\Rightarrow\frac{3}{5}\times y=\frac{113}{60}\)
\(\Rightarrow y=\frac{113}{36}\)
\(\frac{1}{2}:\frac{3}{2}:\frac{5}{4}:\frac{6}{5}:\frac{7}{6}:\frac{8}{7}\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\)
\(=\frac{1\cdot\left(2\cdot5\cdot6\cdot7\right)}{8\cdot3\cdot\left(2\cdot5\cdot6\cdot7\right)}\)
\(=\frac{1}{24}\)
\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\cdot\frac{8}{9}\cdot\frac{9}{10}\)
\(=\frac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)}{\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)\cdot10}\)
\(=\frac{1}{10}\)
Mình làm luôn ko chép lại đề nhé
= ( \(\frac{1}{3}+\frac{3}{3}\)) x ( \(\frac{1}{4}+\frac{4}{4}\)) x ... x ( \(\frac{1}{16}+\frac{16}{16}\))
= \(\frac{4}{3}x\frac{5}{4}x...x\frac{17}{16}\)
= \(\frac{4x5x6x...x17}{3x4x5x...x16}\)
= \(\frac{17}{3}\)
(1-1/2)×(1-1/3)×(1-1/4 )×...×(1-1/2006)×(1-1/2007)
=(2/2-1/2)×(3/3-1/3)×(4/4-1/4)×..×(2006/2006-1/2006)×(2007/2007-1/2007)
=1/2×2/3×3/4×..×2005/2006×2006/2007
=1/2007
\(1,\frac{3}{5}+\frac{7}{2}-\frac{4}{7}=\frac{42}{70}+\frac{245}{70}-\frac{40}{70}=\frac{247}{70}\)
\(2,\frac{4}{5}-\frac{1}{3}+\frac{3}{4}=\frac{48}{60}-\frac{20}{60}+\frac{45}{60}=\frac{73}{60}\)
\(3,\frac{1}{3}\times\frac{5}{7}:\frac{7}{2}=\frac{5}{21}:\frac{7}{2}=\frac{10}{147}\)
\(4,\frac{1}{2}:\frac{2}{7}\times\frac{4}{7}=\frac{7}{4}\times\frac{4}{7}=1\)
\(5,\frac{19}{15}\times\frac{3}{7}:\frac{4}{9}=\frac{19}{35}:\frac{4}{9}=\frac{171}{140}\)
\(6,\frac{11}{3}:\frac{9}{16}:\frac{3}{8}=\frac{176}{27}:\frac{3}{8}=\frac{1408}{81}\)
_Chúc bạn học tốt _
B=1/2x2/3x3/4x4/5+....x2003/2004
=1x2x3x4x...x2003/2x3x4x5x...x2004
=1/2004
B=1/2.2/3.3/4.4/5.....2003/2004
B=1/2004
(trên tử có 1.2.3.4...2003,mk bớt ở dưới mẫu =trên tử thì sẽ đc kq như z)
(bn viết ra sẽ rõ)