Tính:
\(\dfrac{-1}{39}+\dfrac{-1}{52}\)
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\(\dfrac{\dfrac{1}{6}-\dfrac{1}{39}+\dfrac{1}{51}}{\dfrac{1}{8}-\dfrac{1}{52}+\dfrac{1}{68}}\)
\(\dfrac{11}{\dfrac{78}{\dfrac{11}{104}+\dfrac{1}{68}}}+\dfrac{1}{51}\)
\(\dfrac{71}{\dfrac{442}{\dfrac{213}{1768}}}\)\(\Rightarrow\dfrac{4}{3}\)
Bài 1:
\(\dfrac{-13}{38}\) và \(\dfrac{29}{-88}\)
\(\dfrac{-13}{38}=\dfrac{-13.29}{38.29}=\dfrac{-377}{1102}\)
\(\dfrac{29}{-88}=\dfrac{-29}{88}=\dfrac{-29.13}{88.13}=\dfrac{-377}{1144}\)
Vì \(\dfrac{-377}{1102}< \dfrac{-377}{1144}\) nên \(\dfrac{-13}{38}< \dfrac{29}{-88}\)
\(\dfrac{-18}{31}\) và \(\dfrac{-1818}{3131}\)
\(\dfrac{-18}{31}\)
\(\dfrac{-1818}{3131}=\dfrac{-1818:101}{3131:101}=\dfrac{-18}{31}\)
Vì \(\dfrac{-18}{31}=\dfrac{-18}{31}\) nên \(\dfrac{-18}{31}=\dfrac{-1818}{3131}\)
Bài 2:
a) \(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-4+-3}{156}=\dfrac{-7}{156}\)
b) \(\dfrac{-6}{9}+\dfrac{-12}{16}=\dfrac{-2}{3}+\dfrac{-3}{4}=\dfrac{-8}{12}+\dfrac{-9}{12}=\dfrac{-17}{12}\)
\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-52}{2028}+\dfrac{-39}{2028}=\dfrac{\left(-52+-39\right)}{2028}=\dfrac{-91}{2028}=\dfrac{-7}{156}\)
Chúc bạn học tốt!
a: \(A=\left[6\cdot\dfrac{1}{27}+3\cdot\dfrac{1}{3}+1\right]:\dfrac{-4}{3}\)
\(=\left(\dfrac{2}{9}+2\right)\cdot\dfrac{-3}{4}\)
\(=\dfrac{20}{9}\cdot\dfrac{-3}{4}=\dfrac{-60}{36}=\dfrac{-5}{3}\)
b: \(B=\dfrac{\dfrac{1}{3}\left(\dfrac{1}{13}-\dfrac{1}{2}-\dfrac{1}{17}\right)}{-\dfrac{1}{4}\left(\dfrac{1}{13}-\dfrac{1}{2}-\dfrac{1}{17}\right)}:\dfrac{11}{6}\)
\(=\dfrac{-1}{3}:\dfrac{1}{4}\cdot\dfrac{6}{11}=\dfrac{-4}{3}\cdot\dfrac{6}{11}=\dfrac{-24}{33}=\dfrac{-8}{11}\)
a) \(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-7}{156}\)
b) \(\dfrac{-6}{9}+\dfrac{-12}{16}=\dfrac{-2}{3}+\dfrac{-3}{4}=\dfrac{-8}{12}+\dfrac{-9}{12}=\dfrac{-17}{12}\)
c) \(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-22}{55}-\dfrac{-15}{55}=\dfrac{-7}{55}\)
d) \(\dfrac{-34}{37}.\dfrac{74}{-85}=\dfrac{4}{5}\)
e) \(\dfrac{-5}{9}:\dfrac{-7}{18}=\dfrac{-5}{9}.\dfrac{18}{-7}=\dfrac{10}{7}\)
Ta có: \(\dfrac{-1}{3}\cdot\dfrac{141}{17}-\dfrac{39}{3}\cdot\dfrac{-1}{17}\)
\(=\dfrac{-1}{3}\cdot\dfrac{141}{17}-\dfrac{-1}{3}\cdot\dfrac{39}{17}\)
\(=\dfrac{-1}{3}\cdot\left(\dfrac{141}{17}-\dfrac{39}{17}\right)\)
\(=\dfrac{-1}{3}\cdot6=-2\)
\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4-3}{156}=-\dfrac{7}{156}\)
\(\dfrac{-52}{2028}+\dfrac{-39}{2028}=\dfrac{-7}{156}\)