Tính \(A=\frac{8\frac{3}{9}.5\frac{1}{4}+3\frac{16}{19}.5\frac{1}{4}}{\left(2\frac{14}{17}-2\frac{1}{34}\right)34};\frac{7}{24}\)
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MT
5 tháng 10 2019
\(A=\frac{8\frac{3}{9}.5\frac{1}{4}+3\frac{16}{19}.5\frac{1}{4}}{\left(2\frac{14}{17}-2\frac{1}{34}\right).34}:\frac{7}{24}\)
\(=\frac{5\frac{1}{4}\left(8\frac{1}{3}+3\frac{16}{19}\right)}{\left(\frac{28}{34}-\frac{1}{34}\right).34}:\frac{7}{24}\)
\(=\frac{\frac{21}{4}\left(\frac{25}{3}+\frac{73}{19}\right)}{\frac{27}{34}.34}:\frac{7}{24}\)
\(=\frac{\frac{21}{4}\left(\frac{475}{57}+\frac{219}{57}\right)}{27}:\frac{7}{24}\)
\(=\frac{\frac{21}{4}.\frac{674}{57}}{27}:\frac{7}{24}\)
\(=\frac{\frac{14154}{228}}{27}.\frac{24}{7}=\frac{\frac{339696}{228}}{189}\)
15 tháng 2 2020
\(A=\frac{8\frac{3}{9}\cdot5\frac{1}{4}+3\frac{16}{19}\cdot5\frac{1}{4}}{\left(2\frac{14}{17}-2\frac{1}{34}\right)\cdot34}:\frac{7}{24}\)
\(A=\frac{\left(8\frac{3}{9}+3\frac{16}{19}\right)\cdot5\frac{1}{4}}{\left(2\frac{14}{17}-2\frac{1}{34}\right)\cdot34}\cdot\frac{24}{7}\)
\(A=\frac{\left(\frac{25}{3}+\frac{73}{19}\right)\cdot\frac{21}{4}}{\left(\frac{48}{17}-\frac{69}{34}\right)\cdot34}\cdot\frac{24}{7}\)
\(A=\frac{\frac{694}{57}\cdot\frac{21}{4}}{\frac{27}{34}\cdot34}\cdot\frac{24}{7}=\frac{\left(\frac{2429}{38}\right)}{27}\cdot\frac{24}{7}=\frac{1388}{171}\)
P/S : Số lớn quá
1 tháng 2 2020
\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)
28 tháng 2 2022
a: \(A=\left(\dfrac{15}{34}+\dfrac{9}{34}-1-\dfrac{15}{17}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\left(\dfrac{12}{17}-1-\dfrac{15}{17}\right)+1\)
\(=\dfrac{-20}{17}+1=\dfrac{-3}{17}\)
b: \(B=\dfrac{-5}{3}\cdot16\dfrac{2}{7}-\dfrac{-5}{3}\cdot28\dfrac{2}{7}\)
\(=\dfrac{-5}{3}\left(16+\dfrac{2}{7}-28-\dfrac{2}{7}\right)=\dfrac{-5}{3}\cdot\left(-12\right)=20\)
c: \(C=25\cdot\dfrac{-1}{27}+\dfrac{1}{5}-2\cdot\dfrac{1}{4}-\dfrac{1}{2}\)
\(=\dfrac{-25}{27}+\dfrac{1}{5}-1\)
\(=\dfrac{-125+27-135}{135}=\dfrac{-233}{135}\)
