1) Tính:
a) \(\frac{2}{3}-4.\left(\frac{1}{2}+\frac{3}{4}\right)\)
b)\(\left(-\frac{1}{3}+\frac{5}{6}\right).11-7\)
c) \(\frac{3}{11}.\frac{4}{15}+\frac{19}{15}.\frac{-3}{11}\)
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22 tháng 3 2020
\(A=\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{6-\frac{43}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\frac{50}{31}\cdot\frac{31}{50}=1\)
18 tháng 8 2020
[\(\frac{-75}{59}\).\(\frac{-107}{93}\)]\(\frac{31}{50}\)=\(\frac{2675}{1829}\).\(\frac{31}{50}\)=\(\frac{107}{118}\)
18 tháng 8 2020
\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{1}{6}\cdot\frac{20}{3}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{6-15+\frac{2}{3}}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\left(-\frac{150}{107}\right)\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}=\frac{50}{31}\cdot\frac{31}{50}=1\)
a/ \(\frac{2}{3}-4.\left(\frac{1}{2}+\frac{3}{4}\right)\)
= \(\frac{2}{3}-2-3\)
=\(\frac{2}{3}-5\)
=\(\frac{2}{3}-\frac{15}{3}\)
=\(\frac{-13}{3}\)
b/\(\left(-\frac{1}{3}+\frac{5}{6}\right).11-7\)
=\(\frac{-11}{3}+\frac{55}{6}-7\)
=\(\frac{-22}{6}+\frac{55}{6}-7\)
= \(\frac{11}{2}-\frac{14}{2}\)
=\(\frac{-3}{2}\)
c/\(\frac{3}{11}.\frac{4}{15}+\frac{19}{15}.\frac{-3}{11}\)
=\(\frac{4}{55}\)\(-\frac{19}{55}\)
=\(\frac{-3}{11}\)
a) \(\frac{2}{3}-4.\left(\frac{1}{2}+\frac{3}{4}\right)=\frac{2}{3}-4.\left(\frac{2}{4}+\frac{3}{4}\right)\)
\(=\frac{2}{3}-4.\frac{5}{4}=\frac{2}{3}-5=\frac{-13}{3}\)
b) \(\left(-\frac{1}{3}+\frac{5}{6}\right).11-7\)
\(=\frac{3}{6}.11-7=\frac{1}{2}.11-7\)
\(=\frac{11}{2}-7=-\frac{3}{2}\)
c) \(\frac{3}{11}.\frac{4}{15}+\frac{19}{15}.\frac{-3}{11}\)
\(=\frac{3}{11}.\left(\frac{4}{15}-\frac{19}{15}\right)\)
\(=\frac{3}{11}.\left(-1\right)=-\frac{3}{11}\)
Chúc bạn học tốt nhé !