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\(\dfrac{5}{6}\cdot\dfrac{1}{3}+\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{3}\)
\(=\dfrac{5\cdot1}{6\cdot3}+\dfrac{1}{2}\cdot\dfrac{3\cdot5}{4\cdot3}\)
\(=\dfrac{5}{18}+\dfrac{1}{2}\cdot\dfrac{15}{12}\)
\(=\dfrac{5}{18}+\dfrac{1}{2}\cdot\dfrac{5}{4}\)
\(=\dfrac{5}{18}+\dfrac{1\cdot5}{2\cdot8}\)
\(=\dfrac{5}{18}+\dfrac{5}{8}\)
\(=\dfrac{65}{72}\)
\(13,=-1-1-...-1+101\)
Tổng trên có \(\left(100-1+1\right):2=50\) (số -1)
Vậy tổng trên \(=-50+101=51\)
a: \(=\dfrac{-5}{7}-\dfrac{2}{7}+\dfrac{3}{4}+\dfrac{1}{4}-\dfrac{1}{5}=-\dfrac{1}{5}\)
b: \(=\dfrac{-3}{31}-\dfrac{28}{31}+\dfrac{-6}{17}-\dfrac{11}{17}+\dfrac{1}{29}-\dfrac{1}{5}=\dfrac{-24}{145}\)
1+2+3+4+5+6+7+8
= (1+8) x 4 (8 số hạng chia thành 4 cặp)
= 9 x 4
= 36
\(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{-5}{7}+\dfrac{1}{6}-\dfrac{3}{35}+\dfrac{1}{3}-\dfrac{-1}{41}\)
\(=\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(-\dfrac{1}{5}-\dfrac{5}{7}-\dfrac{3}{35}\right)+\dfrac{1}{41}\)
\(=\dfrac{3+2+1}{6}+\dfrac{-7-25-3}{35}+\dfrac{1}{41}\)
\(=\dfrac{6}{6}+\dfrac{-35}{35}+\dfrac{1}{41}=\dfrac{1}{41}\)
\(\dfrac{5}{6}.\dfrac{1}{3}+\dfrac{5}{6}+\dfrac{3}{4}\)
\(=\dfrac{5}{6}.\left(\dfrac{1}{3}+\dfrac{3}{4}\right)=\dfrac{5}{6}.\dfrac{7}{12}=\dfrac{35}{72}\)