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\(B=\dfrac{13}{15}\cdot0,15\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):\left(1\dfrac{23}{24}\right)\)
\(=\dfrac{13}{15}\cdot\dfrac{15}{100}\cdot3+\left(\dfrac{32}{60}-1-\dfrac{19}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{39}{100}+\dfrac{-47}{60}\cdot\dfrac{24}{47}\)
\(=\dfrac{39}{100}-\dfrac{24}{60}=\dfrac{39}{100}-\dfrac{40}{100}=-\dfrac{1}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 ...
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1\dfrac{13}{15}\cdot\left(0,5\right)^2+3\cdot\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
\(=\dfrac{28}{15}\cdot\dfrac{1}{4}+3\cdot\left(\dfrac{8}{15}+\dfrac{79}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{7}{15}+3\cdot\dfrac{37}{20}\cdot\dfrac{24}{47}\)
\(=\dfrac{7}{15}+\dfrac{666}{235}=\dfrac{2327}{705}\)
\(1\dfrac{13}{15}.\left(0,5\right)^2+3.\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
\(=\dfrac{28}{15}.\dfrac{1}{4}+3.\left(\dfrac{8}{15}+\dfrac{79}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{7}{15}+3.\dfrac{37}{20}:\dfrac{47}{24}\)
\(=\dfrac{7}{15}+\dfrac{666}{235}\)
\(=\dfrac{2327}{705}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\\ =\dfrac{28}{15}.0,25.3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\\ =\dfrac{7}{15}.3+\dfrac{-47}{60}:\dfrac{47}{24}\\ =\dfrac{7}{5}+\dfrac{-2}{5}\\ =\dfrac{5}{5}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)