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Giải:
a) \(2\dfrac{17}{20}-1\dfrac{15}{11}+6\dfrac{9}{20}:3\)
\(=\dfrac{57}{20}-\dfrac{26}{11}+\dfrac{129}{20}:3\)
\(=\dfrac{107}{220}+\dfrac{43}{20}\)
\(=\dfrac{29}{11}\)
b) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\)
\(=\dfrac{31}{7}:\left(\dfrac{7}{5}.\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}\)
\(=\dfrac{5}{7}\)
c) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\)
\(=\left(\dfrac{29}{9}.\dfrac{15}{23}.\dfrac{36}{29}\right):\dfrac{5}{23}\)
\(=\dfrac{60}{23}:\dfrac{5}{23}\)
\(=12\)
\(a,11\dfrac{3}{4}-\left(6\dfrac{5}{6}-4\dfrac{1}{2}\right)+1\dfrac{2}{3}\)
\(=\dfrac{47}{4}-\left(\dfrac{41}{6}-\dfrac{9}{2}\right)+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\left(\dfrac{41}{6}-\dfrac{27}{6}\right)+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\dfrac{14}{6}+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\dfrac{7}{3}+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\left(\dfrac{7}{3}+\dfrac{5}{3}\right)\)
\(=\dfrac{47}{4}-\dfrac{12}{3}\)
\(=\dfrac{47}{4}-4\)
\(=\dfrac{47}{4}-\dfrac{16}{4}\)
\(=\dfrac{31}{4}\)
c) Ta có: \(4\dfrac{3}{7}:\left(\dfrac{7}{5}\cdot4\dfrac{3}{7}\right)\)
\(=\dfrac{31}{7}:\left(\dfrac{7}{5}\cdot\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}\)
\(=\dfrac{5}{7}\)
a: =11+3/4-6-5/6+4+1/2+1+2/3
=10+9/12-10/12+6/12+8/12
=10+13/12=133/12
b: \(=2+\dfrac{17}{20}-1-\dfrac{11}{15}+2+\dfrac{3}{20}\)
=3-11/15
=34/15
c: \(=\dfrac{31}{7}:\left(\dfrac{7}{5}\cdot\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}=\dfrac{5}{7}\)
d: \(=\dfrac{29}{8}\cdot\dfrac{36}{29}\cdot\dfrac{15}{23}\cdot\dfrac{23}{5}=\dfrac{9}{2}\cdot3=\dfrac{27}{2}\)
