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Bài 1:
a) Ta có: \(\frac{3}{5}+\frac{4}{15}\)
\(=\frac{9}{15}+\frac{4}{15}\)
\(=\frac{13}{15}\)
b) Ta có: \(\frac{-3}{5}+\frac{5}{7}\)
\(=\frac{-21}{35}+\frac{25}{35}=\frac{4}{35}\)
c) Ta có: \(\frac{5}{6}:\frac{-7}{12}\)
\(=\frac{5}{6}\cdot\frac{-12}{7}=\frac{-60}{42}=\frac{-10}{7}\)
d) Ta có: \(\frac{-21}{24}:\frac{-14}{8}\)
\(=\frac{-7}{8}:\frac{-7}{4}\)
\(=\frac{-7}{8}\cdot\frac{4}{-7}=\frac{4}{8}=\frac{1}{2}\)
e) Ta có: \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}\)
\(=\frac{-3}{5}\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)\)
\(=-\frac{3}{5}\cdot2=\frac{-6}{5}\)
f) Ta có: \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{3}\)
\(=\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{1}{3}\cdot4\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}-4\right)\)
\(=\frac{1}{3}\cdot\left(-2\right)=\frac{-2}{3}\)
g) Ta có: \(\frac{4}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{19}+\frac{5}{7}\)
\(=\frac{4}{19}\cdot\frac{-3}{7}+\frac{5}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{-3}\)
\(=-\frac{3}{7}\left(\frac{4}{19}+\frac{5}{19}+\frac{-5}{3}\right)\)
\(=\frac{-3}{7}\cdot\left(\frac{27}{57}+\frac{-95}{57}\right)\)
\(=\frac{-3}{7}\cdot\frac{-68}{57}=\frac{68}{133}\)
h) Ta có: \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}\)
\(=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{5}{13}\right)\)
\(=\frac{5}{9}\)
e. \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}=-\frac{3}{5}\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)=-\frac{3}{5}\cdot2=-\frac{6}{5}\)
f. \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{5}=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}\right)-\frac{4}{5}=\frac{1}{3}\cdot2-\frac{4}{5}=\frac{2}{3}-\frac{4}{5}=-\frac{2}{15}\)
g. \(\frac{4}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{15}{19}+\frac{5}{7}=-\frac{3}{7}\left(\frac{4}{19}+\frac{15}{19}\right)+\frac{5}{7}=-\frac{3}{7}\cdot1+\frac{5}{7}=-\frac{3}{7}+\frac{5}{7}=\frac{2}{7}\)
h. \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\cdot1=\frac{5}{9}\)
câu a
\(A=\frac{33.10^3}{2^3.5.10^3+7000}=\frac{33.10^3}{2^3.5.10^3+7.10^3}=\frac{33.10^3}{10^3\left(2^3.5+7\right)}=\frac{33.10^3}{10^3.47}=\frac{33}{47}\)
\(B=\frac{3774}{5217}=\frac{34.111}{47.111}=\frac{34}{47}\)
\(\Rightarrow\frac{33}{47}< \frac{34}{47}\)
=> A<B