a: \(=\dfrac{7+3}{6}\cdot\dfrac{1}{2}-2:\dfrac{7+3}{6}\)

\(=\dfrac{10}{12}-2\cdot\dfrac{6}{10}\)

\(=\dfrac{5}{6}-\dfrac{6}{5}=\dfrac{25-36}{30}=-\dfrac{11}{30}\)

b: \(=\left|\dfrac{10}{5}-\dfrac{2}{5}\right|\cdot\dfrac{1}{27}-\dfrac{3}{5}+1\)

\(=\dfrac{8}{5}\cdot\dfrac{1}{27}-\dfrac{3}{5}+1\)

\(=\dfrac{8}{135}-\dfrac{81}{135}+\dfrac{135}{135}=\dfrac{62}{135}\)

\(a)2:\left(\dfrac{1}{2}-\dfrac{2}{3}\right)^3=2:\left(\dfrac{-1}{6}\right)^3=2:\dfrac{-1}{216}=2.\left(-216\right)=-432\)

\(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)

\(=\left(\dfrac{1}{3}-\dfrac{1}{4}\right).\left(\dfrac{1}{20}\right)^2\)

\(=\dfrac{1}{12}.\dfrac{1}{400}=\dfrac{1}{4800}\)

chúc bạn học tốt

\(=\left(\dfrac{88-33+60}{132}\right):\dfrac{55+132-84}{132}\)

\(=\dfrac{115}{103}\)

\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{264}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{1}{2}\cdot\dfrac{3}{4}+\dfrac{5}{8}=\dfrac{3}{8}+\dfrac{5}{8}=1\)

\(\dfrac{1}{A}=\dfrac{\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)-\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{4}}{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}\)

\(\dfrac{1}{A}=1-\dfrac{\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{4}}{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}\)

\(\dfrac{1}{A}=1-\dfrac{\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{4}}{\dfrac{3.4+2.4-2.3}{2.3.4}}\)

\(\dfrac{1}{A}=\dfrac{1}{3.4+2.4-2.3}\)

\(\dfrac{1}{A}=1-\dfrac{1}{14}\) \(=\dfrac{13}{14}\)

⇒ \(A=\dfrac{14}{13}\)

Cách 2:

( 1 )

Có: \(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\)\(=\dfrac{12}{24}+\dfrac{8}{24}-\dfrac{6}{24}=\dfrac{14}{24}\)

Thay \(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\) \(=\dfrac{14}{24}\) vào ( 1 ), ta có:

\(\dfrac{\dfrac{14}{24}}{\dfrac{14}{24}-\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{4}}\) \(=\dfrac{\dfrac{14}{24}}{\dfrac{14}{24}-\dfrac{1}{24}}\) \(=\dfrac{\dfrac{14}{24}}{\dfrac{13}{24}}\) \(=\dfrac{14}{24}:\dfrac{13}{24}=\dfrac{14.24}{13.24}=\dfrac{14}{13}\)

Vậy \(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}{\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)-\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{4}}\) \(=\dfrac{14}{13}\).

a: \(1\dfrac{1}{2}\cdot1\dfrac{1}{3}\cdot1\dfrac{1}{4}\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\)

\(=\dfrac{5}{2}\)

b: \(1\dfrac{1}{2}:1\dfrac{1}{3}:1\dfrac{1}{4}\)

\(=\dfrac{3}{2}:\dfrac{4}{3}:\dfrac{5}{4}\)

\(=\dfrac{3}{2}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}=\dfrac{9}{10}\)

\(A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{64}\)

\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{64}\)

=1/64

a) \(\dfrac{-5}{9}.\dfrac{3}{11}+\dfrac{-13}{18}.\dfrac{3}{11}\)

\(=\dfrac{3}{11}.\left(\dfrac{-5}{9}+\dfrac{-13}{9}\right)\)

\(=\dfrac{3}{11}.\left(-2\right)\)

\(=\dfrac{-6}{11}\)

b) \(\dfrac{11}{2}.2\dfrac{1}{3}-1\dfrac{1}{5}.1\dfrac{1}{2}\)

\(=\dfrac{11}{3}.\dfrac{7}{3}-\dfrac{6}{5}.\dfrac{3}{2}\)

\(=\dfrac{77}{9}-\dfrac{9}{5}\)

\(=\dfrac{385}{45}-\dfrac{81}{45}\)

\(=\dfrac{304}{45}\)

c) \(1\dfrac{1}{9}.\dfrac{2}{145}-4\dfrac{1}{3}-\dfrac{2}{145}+\dfrac{2}{145}\)

\(=\dfrac{10}{9}.\dfrac{2}{145}-\dfrac{8}{3}\)

\(=\dfrac{4}{261}-\dfrac{8}{3}\)

\(=\dfrac{4}{261}-\dfrac{696}{261}\)

\(=-\dfrac{692}{261}\)

d) \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)

\(=\left(1-1\right)+\left(2-2\right)+\left(3-3\right)+4-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)\)

\(=0+0+0+4-1-1-1\)

\(=4-3\)

\(=1\)