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2 tháng 10 2021
a: \(\left(0.5\right)^3\cdot2^3=1\)
b: \(\left(0.25\right)^2\cdot16=1\)
c: \(\left(\dfrac{3}{5}\right)^3:\left(-\dfrac{27}{1000}\right)=\dfrac{3^3}{5^3}\cdot\dfrac{-1000}{27}=\dfrac{-1000}{125}=-8\)
Y
1
4 tháng 10 2021
\(=\dfrac{2^4\cdot5^4+2^5\cdot5^3}{2^{10}\cdot16}=\dfrac{2^4\cdot5^3\left(5+2\right)}{2^{10}\cdot2^4}=\dfrac{2^4\cdot5^3\cdot7}{2^{14}}=\dfrac{5^3\cdot7}{2^{10}}=\dfrac{875}{1024}\)
Y
15 tháng 8 2023
\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)
Vậy \(A=\dfrac{1}{2}\)
\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)
Vậy \(B=3\)
17 tháng 9 2023
\(a.\left[-\dfrac{6}{11}.\dfrac{11}{-6}\right].\dfrac{7}{10}.\left(-20\right)=1.7.\left(-2\right)=-14\)
\(b.\dfrac{-1}{2}:\dfrac{3}{4}.\dfrac{-7}{2}=\dfrac{7}{4}:\dfrac{3}{4}=\dfrac{7}{3}\)
\(c.\dfrac{93}{7}:-\dfrac{8}{9}+\dfrac{19}{7}:\dfrac{-8}{9}=\left(\dfrac{93}{7}+\dfrac{19}{7}\right):-\dfrac{8}{9}=\dfrac{-9}{8}.\dfrac{112}{7}=-18\)
PT
0
18 tháng 9 2023
a) \(=\left(13\dfrac{2}{7}+2\dfrac{5}{7}\right):\left(-\dfrac{8}{9}\right)\)
\(=16:\dfrac{-8}{9}=\dfrac{-8\cdot\left(-2\right)\cdot9}{-8}=-18\)
b)
\(=\left(\dfrac{-6}{11}\cdot\dfrac{11}{-6}\right)\cdot\dfrac{7\cdot10\cdot\left(-2\right)}{10}\)
\(=-14\)
c) \(=\dfrac{-1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{-7}{2}\)
\(=\dfrac{-1\cdot2\cdot2\cdot\left(-7\right)}{2\cdot3\cdot2}=\dfrac{7}{3}\)
S
24 tháng 12 2022
\(\left(0.25\right)^{10}.4^{10}+\sqrt{5^2-3^2}\)
\(=0.4^{10}+\sqrt{25-9}\)
\(=0+\sqrt{16}=0+4=4\)
\(\dfrac{5}{20}+\dfrac{18}{11}-25\%-\left(\dfrac{18}{11}-\dfrac{4}{9}\right)\)
\(=\dfrac{5}{20}+\dfrac{18}{11}-\dfrac{1}{4}-\dfrac{18}{11}+\dfrac{4}{9}\)
\(=\left(\dfrac{5}{20}-\dfrac{1}{4}\right)+\left(\dfrac{18}{11}-\dfrac{18}{11}\right)+\dfrac{4}{9}\)
\(=0+0+\dfrac{4}{9}=\dfrac{4}{9}\)
TP
2
20 tháng 8 2018
\(A=\left(\frac{5}{3}-\frac{3}{7}+9\right)-\left(2+\frac{5}{7}-\frac{2}{3}\right)+\left(\frac{8}{7}-\frac{4}{3}-10\right)\)
\(=\frac{5}{3}-\frac{3}{7}+9-2-\frac{5}{7}+\frac{2}{3}+\frac{8}{7}-\frac{4}{3}-10\)
\(=\left(\frac{5}{3}+\frac{2}{3}-\frac{4}{3}\right)-\left(\frac{3}{7}+\frac{5}{7}-\frac{8}{7}\right)+\left(9-2-10\right)\)
\(=1-0-3\)
\(=-2\)
20 tháng 8 2018
5/3-3/7+9-2-5/7+2/3+8/7-4/3-10
(5/3-4/3+2/3)+(8/7-3/7-5/7)+(9-2-10)
1+0-3=-2
MD
2
18 tháng 10 2017
\(\left(\frac{3}{5}\right)^{10}.\left(\frac{5}{3}\right)^{10}-\frac{13^4}{39^4}+2017^0\\ =\left(\frac{3}{5}.\frac{5}{3}\right)^{10}-\left(\frac{13}{39}\right)^4+1\\ =1^{10}-\left(\frac{1}{3}\right)^4+1\)
\(=1-\frac{1}{81}+1\\ =2+\frac{-1}{81}\\ =\frac{161}{81}\)
\(\dfrac{4^5.21+4^5.19}{2^{10}.11+2^{10}.5}=\dfrac{4^5.\left(21+19\right)}{2^{10}.\left(11+5\right)}=\dfrac{4^5.30}{2^{10}.16}\)\(=\dfrac{\left(2.2\right)^5.15}{2^{10}.8}=\dfrac{2^5.2^5.15}{2^{10}.8}=\dfrac{15}{8}\)