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Giải:
a) \(11\dfrac{3}{4}.\left(6\dfrac{5}{6}-4\dfrac{1}{2}+1\dfrac{2}{3}\right)\)
\(=\dfrac{47}{4}.\left(\dfrac{41}{6}-\dfrac{9}{2}+\dfrac{5}{3}\right)\)
\(=\dfrac{47}{4}.4\)
\(=47\)
b) \(\left(5\dfrac{7}{8}-2\dfrac{1}{4}-0,5\right):2\dfrac{23}{26}\)
\(=\left(\dfrac{47}{8}-\dfrac{9}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}\)
\(=\dfrac{25}{8}:\dfrac{75}{26}\)
\(=\dfrac{13}{12}\)
c) \(\left(17\dfrac{13}{15}-3\dfrac{3}{7}\right)-\left(2\dfrac{12}{15}-4\right)\)
\(=\dfrac{268}{15}-\dfrac{24}{7}-\dfrac{14}{5}+4\)
\(=\left(\dfrac{268}{15}-\dfrac{14}{5}\right)+\left(\dfrac{-24}{7}+4\right)\)
\(=\dfrac{226}{15}+\dfrac{4}{7}\)
\(=\dfrac{1642}{105}\)
d) \(2\dfrac{2}{3}.\left(\dfrac{-4}{5}.0,375.-10.\dfrac{-15}{24}\right)\)
\(=\dfrac{8}{3}.\left(\dfrac{-4}{5}.\dfrac{3}{8}.-10.\dfrac{-5}{8}\right)\)
\(=\left(\dfrac{8}{3}.\dfrac{3}{8}\right).\left(\dfrac{-4}{5}.\dfrac{-5}{8}.-10\right)\)
\(=1.-5\)
\(=-5\)
Chúc bạn học tốt!
Bài 1 :
\(a)\frac{-17}{30}-\frac{11}{-15}+\left(-\frac{7}{12}\right)\)
\(=\frac{1}{6}+\left(-\frac{7}{12}\right)\)
\(=-\frac{5}{12}\)
\(b)-\frac{5}{9}+\frac{5}{9}:\left(1\frac{2}{3}-2\frac{1}{12}\right)\)
\(=-\frac{5}{9}+\frac{5}{9}:\left(\frac{5}{3}-\frac{25}{12}\right)\)
\(=-\frac{5}{9}+\frac{5}{9}:\left(-\frac{5}{12}\right)\)
\(=-\frac{5}{9}+\left(-\frac{2}{3}\right)\)
\(=-\frac{1}{9}\)
\(c)-\frac{7}{25}\times\frac{11}{13}+\left(-\frac{7}{25}\right)\times\frac{2}{13}-\frac{18}{25}\)
\(=-\frac{77}{325}+\left(-\frac{14}{325}\right)-\frac{18}{25}\)
\(=-\frac{7}{25}-\frac{18}{25}\)
\(=-1\)
a) Ta có:
\(A=-3\cdot7\cdot\left(-2\right)\cdot\left(-13\right)\)
\(A=-21\cdot26\)
\(A=-546\)
\(B=-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot\left(-4\right)\cdot5\)
\(B=2\cdot12\cdot5\)
\(B=2\cdot60\)
\(B=120\)
Mà: \(120>-546\)
\(\Rightarrow B>A\)
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: \(=\dfrac{29}{8}\cdot\dfrac{36}{29}\cdot\dfrac{15}{23}\cdot\dfrac{23}{5}=\dfrac{9}{2}\cdot3=\dfrac{27}{2}\)
\(a.\frac{8}{7}+\frac{4}{7}\times\left(-\frac{6}{11}\right)-\frac{4}{7}\times\frac{5}{11}\)
\(=\frac{8}{7}+\frac{4}{7}\left(-\frac{6}{11}-\frac{5}{11}\right)\)
\(=\frac{8}{7}+\frac{4}{7}.\left(-1\right)\)
\(=\frac{8}{7}-\frac{4}{7}\)
\(=\frac{4}{7}\)
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
\(a,11\dfrac{3}{4}-\left(6\dfrac{5}{6}-4\dfrac{1}{2}\right)+1\dfrac{2}{3}\)
\(=\dfrac{47}{4}-\left(\dfrac{41}{6}-\dfrac{9}{2}\right)+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\left(\dfrac{41}{6}-\dfrac{27}{6}\right)+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\dfrac{14}{6}+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\dfrac{7}{3}+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\left(\dfrac{7}{3}+\dfrac{5}{3}\right)\)
\(=\dfrac{47}{4}-\dfrac{12}{3}\)
\(=\dfrac{47}{4}-4\)
\(=\dfrac{47}{4}-\dfrac{16}{4}\)
\(=\dfrac{31}{4}\)
c) Ta có: \(4\dfrac{3}{7}:\left(\dfrac{7}{5}\cdot4\dfrac{3}{7}\right)\)
\(=\dfrac{31}{7}:\left(\dfrac{7}{5}\cdot\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}\)
\(=\dfrac{5}{7}\)