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88986 x 2003 - 678
= ( 88968 x 2003 ) - 678
= 178238958 - 578
= 178238280
Sao giống bài lớp 7 quá he
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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) \(\frac{3}{5}\times y-\frac{4}{5}:3+\frac{1}{12}=\frac{3}{2}+\frac{1}{5}\)
\(\Rightarrow\frac{3}{5}\times y-\frac{4}{15}+\frac{1}{12}=\frac{17}{10}\)
\(\Rightarrow\frac{3}{5}\times y-\frac{4}{15}=\frac{97}{60}\)
\(\Rightarrow\frac{3}{5}\times y=\frac{113}{60}\)
\(\Rightarrow y=\frac{113}{36}\)
\(\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{5}\right)\times\left(1-\frac{1}{6}\right)\times\left(1-\frac{1}{7}\right)\times\left(1-\frac{1}{8}\right)-\frac{1}{4}\times\frac{1}{2}\)
\(=\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}\times\frac{6}{7}\times\frac{7}{8}-\frac{1}{4}\times\frac{1}{2}\)
\(=\frac{2}{8}-\frac{1}{4}\times\frac{1}{2}\)
\(=\frac{2}{8}-\frac{1}{8}=\frac{1}{8}\)
\(\frac{1}{2}:\frac{3}{2}:\frac{5}{4}:\frac{6}{5}:\frac{7}{6}:\frac{8}{7}\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\)
\(=\frac{1\cdot\left(2\cdot5\cdot6\cdot7\right)}{8\cdot3\cdot\left(2\cdot5\cdot6\cdot7\right)}\)
\(=\frac{1}{24}\)
\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\cdot\frac{8}{9}\cdot\frac{9}{10}\)
\(=\frac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)}{\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)\cdot10}\)
\(=\frac{1}{10}\)
\(\left(1-\frac{2}{5}\right)\times\left(1-\frac{2}{7}\right)\times\left(1-\frac{2}{9}\right)\times\left(1-\frac{2}{11}\right)\times\left(1-\frac{2}{13}\right)\)
\(=\frac{3}{5}\times\frac{5}{7}\times\frac{7}{9}\times\frac{9}{11}\times\frac{11}{13}=\frac{3}{13}\)