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Cách 1 ở tất cả các câu: Tính bình thường như đề cho
a) Cách 2: \(\frac{5}{7}\times\frac{3}{4}+\frac{5}{7}\times\frac{1}{2}\)
b) Cách 2: \(\frac{13}{21}\times\left(\frac{5}{7}+\frac{2}{7}\right)\)
c) Cách 2: \(\frac{2}{5}\times\frac{1}{2}+\frac{2}{5}\times\frac{1}{3}\)
d) Cách 2: \(\frac{17}{21}\times\left(5+\frac{2}{5}\right)\)
3+5/21=3/1+5/21=68/21
5-13/7=22/7
5/7×4=20/7
4:1/5=?20
k mk nhé
# chikute #
3+\(\frac{5}{21}\)=\(\frac{68}{21}\)
5-\(\frac{13}{7}\)=\(\frac{22}{7}\)
\(\frac{5}{7}\)*4=\(\frac{20}{7}\)
4:\(\frac{1}{5}\)=20
\(2-\dfrac{5}{7}+\dfrac{9}{21}\)
\(=2-\dfrac{5}{7}+\dfrac{3}{7}\)
\(=\dfrac{14}{7}-\dfrac{5}{7}+\dfrac{3}{7}\)
\(=\dfrac{12}{7}\)
\(2-\dfrac{5}{7}+\dfrac{9}{21}=\dfrac{42}{21}-\dfrac{15}{21}+\dfrac{9}{21}=\dfrac{12}{7}\)
a: =11/4+5/4-9/8
=4-9/8=32/8-9/8=23/8
b: \(=\dfrac{6}{7}\cdot\dfrac{7}{4}+\dfrac{5}{3}=\dfrac{3}{2}+\dfrac{5}{3}=\dfrac{9+10}{6}=\dfrac{19}{6}\)
c: \(=\dfrac{13}{18}\cdot\dfrac{9}{5}-1=\dfrac{13}{10}-1=\dfrac{3}{10}\)
d: \(=3+\dfrac{9}{4}\cdot\dfrac{5}{3}=3+\dfrac{45}{12}=\dfrac{81}{12}=\dfrac{27}{4}\)
\(\dfrac{7}{15}và\dfrac{5}{13}=\dfrac{7.13}{15.13}=\dfrac{91}{195};\dfrac{5}{13}=\dfrac{5.15}{13.15}=\dfrac{75}{195}\)
\(\dfrac{11}{12}và\dfrac{7}{48}=\dfrac{11.4}{12.4}=\dfrac{44}{48};\dfrac{7}{48}\)
\(\dfrac{24}{35}và\dfrac{21}{7}=\dfrac{24}{35};\dfrac{21}{7}=\dfrac{21.5}{7.5}=\dfrac{105}{35}\)
\(\dfrac{15}{16}và\dfrac{7}{8}=\dfrac{15}{16};\dfrac{7}{8}=\dfrac{7.2}{8.2}=\dfrac{14}{16}\)
\(\dfrac{21}{13}:\dfrac{7}{39}-\dfrac{2}{5}=\dfrac{21}{13}\times\dfrac{39}{7}-\dfrac{2}{5}=21\times\dfrac{3}{7}-\dfrac{2}{5}=9-\dfrac{2}{5}=\dfrac{45}{5}-\dfrac{2}{5}=\dfrac{43}{5}\)
\(\dfrac{21}{13}\) : \(\dfrac{7}{39}\) - \(\dfrac{2}{5}\) = 9 - \(\dfrac{2}{5}\) = \(\dfrac{43}{5}\)