Tính:
A = 1 13 15 . 0 , 5 2 . 3 + 8 15 - 1 19 60 : 1 23 24
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\(1\dfrac{13}{15}\cdot\left(0,5\right)^2+3\cdot\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
\(=\dfrac{28}{15}\cdot\dfrac{1}{4}+3\cdot\left(\dfrac{8}{15}+\dfrac{79}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{7}{15}+3\cdot\dfrac{37}{20}\cdot\dfrac{24}{47}\)
\(=\dfrac{7}{15}+\dfrac{666}{235}=\dfrac{2327}{705}\)
\(1\dfrac{13}{15}.\left(0,5\right)^2+3.\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
\(=\dfrac{28}{15}.\dfrac{1}{4}+3.\left(\dfrac{8}{15}+\dfrac{79}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{7}{15}+3.\dfrac{37}{20}:\dfrac{47}{24}\)
\(=\dfrac{7}{15}+\dfrac{666}{235}\)
\(=\dfrac{2327}{705}\)
Giải:
a) \(\left(9\dfrac{4}{9}+5\dfrac{2}{3}\right)-5\dfrac{1}{2}\)
\(=\left(\dfrac{85}{9}+\dfrac{17}{3}\right)-\dfrac{11}{2}\)
\(=\dfrac{136}{9}-\dfrac{11}{2}\)
\(=\dfrac{173}{18}\)
b) \(\dfrac{13}{9}.\dfrac{15}{4}-\dfrac{13}{9}.\dfrac{7}{4}-\dfrac{13}{9}.\dfrac{5}{4}\)
\(=\dfrac{13}{9}.\left(\dfrac{15}{4}-\dfrac{7}{4}-\dfrac{5}{4}\right)\)
\(=\dfrac{13}{9}.\dfrac{3}{4}\)
\(=\dfrac{13}{12}\)
c) \(\dfrac{2}{3}+\dfrac{5}{8}-\dfrac{-1}{3}+0,375\)
\(=\left(\dfrac{2}{3}-\dfrac{-1}{3}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)
\(=1+1\)
\(=2\)
d) \(75\%-3\dfrac{1}{2}+1,5:\dfrac{10}{7}\)
\(=\dfrac{3}{4}+\dfrac{7}{2}+\dfrac{3}{2}:\dfrac{10}{7}\)
\(=\dfrac{3}{4}+\dfrac{7}{2}+\dfrac{21}{20}\)
\(=\dfrac{53}{10}\)
e) \(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
\(=\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{7}{5}+\dfrac{-47}{60}:\dfrac{47}{24}\)
\(=\dfrac{7}{5}+\dfrac{-2}{5}\)
\(=1\)
Giải:
\(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
\(=\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{7}{5}+\left(-\dfrac{47}{60}\right).\dfrac{24}{47}\)
\(=\dfrac{7}{5}+\left(-\dfrac{2}{5}\right)\)
\(=\dfrac{7}{5}-\dfrac{2}{5}\)
\(=1\)
Vậy ...
\(B=\dfrac{13}{15}\cdot0,15\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):\left(1\dfrac{23}{24}\right)\)
\(=\dfrac{13}{15}\cdot\dfrac{15}{100}\cdot3+\left(\dfrac{32}{60}-1-\dfrac{19}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{39}{100}+\dfrac{-47}{60}\cdot\dfrac{24}{47}\)
\(=\dfrac{39}{100}-\dfrac{24}{60}=\dfrac{39}{100}-\dfrac{40}{100}=-\dfrac{1}{100}\)