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A = \(1\dfrac{13}{15}.25.\dfrac{1}{100}.3+\left(\dfrac{8}{15}-\dfrac{78}{60}\right)\div1\dfrac{23}{24}\)
A = \(\dfrac{1.15+13}{15}.25.\dfrac{1}{100}.3+\left(\dfrac{32}{60}-\dfrac{78}{60}\right)\div\dfrac{1.24+23}{24}\)
A = \(\dfrac{28}{15}.25.\dfrac{1}{100}.3+\left(-\dfrac{23}{30}\right)\div\dfrac{47}{24}\)
A = \(\left(\dfrac{28}{15}.25\right).\left(\dfrac{1}{100}.3\right)+\left(-\dfrac{23}{30}\right)\div\dfrac{47}{24}\)
A = \(\left(\dfrac{28.25}{15}\right).\left(\dfrac{1.3}{100}\right)+\left(-\dfrac{23}{30}\right)\div\dfrac{47}{24}\)
A = \(\dfrac{140}{3}.\dfrac{3}{100}+\left(-\dfrac{23}{30}\right)\div\dfrac{47}{24}\)
A = \(\dfrac{7}{5}+\left(-\dfrac{92}{235}\right)=\dfrac{237}{235}\)
\(A=1\dfrac{13}{15}.25.\dfrac{1}{100}.3+\left(\dfrac{8}{15}-\dfrac{78}{60}\right):1\dfrac{23}{24}\\ =\dfrac{28}{15}.25.\dfrac{1}{100}.3-\dfrac{23}{30}:\dfrac{47}{24}\\ =\dfrac{7}{5}-\dfrac{23}{30}.\dfrac{24}{47}\\ =\dfrac{7}{5}-\dfrac{92}{235}\\ =\dfrac{7.47}{235}-\dfrac{92}{235}\\ =\dfrac{237}{235}=1\dfrac{2}{235}\)
a: \(=\dfrac{99}{100}:\left(\dfrac{3}{12}-\dfrac{1}{12}+\dfrac{4}{12}\right)-\dfrac{49}{25}\)
\(=\dfrac{99}{100}:\dfrac{1}{2}-\dfrac{49}{25}\)
\(=\dfrac{99}{50}-\dfrac{98}{50}=\dfrac{1}{50}\)
b: \(=\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{32}{60}-1-\dfrac{19}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{39}{60}+\dfrac{-19}{60}\cdot\dfrac{24}{47}\)
=459/940
Ta có: 1/3 ; 1/15 ; 1/35;...
<=> 1/1.3 ; 1/3.5 ; 1/5.7
=> chữ số thứ 100 là: 1/199.201
Ta có: \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{199.201}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{199}-\frac{1}{201}\)
\(=1-\frac{1}{201}=\frac{200}{201}\)