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NT
1
2 tháng 10 2020
\(a,7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\)
\(=\frac{68}{9}-\frac{11}{4}-\frac{32}{9}\)
\(=4-\frac{11}{4}\)
\(=\frac{5}{4}\)
\(b,\left[\frac{3}{5}.\left(\frac{-7}{10}\right)+\frac{2}{10}.\frac{3}{5}\right]:\left(-3\right)\)
\(=\left(\frac{-21}{50}+\frac{3}{25}\right):\left(-3\right)\)
\(=\frac{-3}{10}:\left(-3\right)\)
\(=\frac{1}{10}\)
NB
1
22 tháng 12 2017
a)7/8*(-5/14)-13/6
=(-5/16)-13/6
=-119/48
b)(-2/3)^2-13/6
=4/9-13/6
=(-31/18)
c)2/3+3/4:(-5/8)
=2/3+(-6/5)
=(-8/15)
28 tháng 6 2016
a) \(\frac{3}{5}\cdot\frac{13}{46}-\frac{1}{10}\cdot\frac{16}{23}=\frac{39-16}{10\cdot23}=\frac{1}{10}\)
b) \(\frac{3}{7}\cdot\frac{9}{26}-\frac{1}{14}\cdot\frac{1}{13}=\frac{27-1}{14\cdot13}=\frac{2\cdot13}{2\cdot7\cdot13}=\frac{1}{7}\)
HN
2
31 tháng 7 2018
\(\left(\frac{2}{3}-\frac{4}{7}\right):\frac{5}{9}+\left(-\frac{8}{7}+\frac{1}{3}\right):\frac{5}{9}\)
\(=\left[\left(\frac{2}{3}-\frac{4}{7}\right)+\left(-\frac{8}{7}+\frac{1}{3}\right)\right]:\frac{5}{9}\)
\(=\left(\frac{2}{3}-\frac{4}{7}-\frac{8}{7}+\frac{1}{3}\right)\cdot\frac{9}{5}\)
\(=\left(1-\frac{12}{7}\right)\cdot\frac{9}{5}\)
\(=-\frac{5}{7}\cdot\frac{9}{5}\)
\(-\frac{9}{7}\)
31 tháng 7 2018
\(\left(\frac{2}{3}-\frac{4}{7}\right):\frac{5}{9}+\left(-\frac{8}{7}+\frac{1}{3}\right):\frac{5}{9}\)
\(=\left(\frac{14}{21}-\frac{12}{21}\right):\frac{5}{9}+\left(-\frac{24}{21}+\frac{7}{21}\right):\frac{5}{9}\)
\(=\frac{2}{21}:\frac{5}{9}+\frac{-17}{21}:\frac{5}{9}\)
\(=\left(\frac{2}{21}+\frac{-17}{21}\right):\frac{5}{9}\)
\(=\frac{-15}{21}:\frac{5}{9}\)
\(=\frac{-15}{21}.\frac{9}{5}\)
\(=\frac{-9}{7}\)
\(=\frac{99}{35}\)
3 tháng 2 2022
a: \(=\dfrac{7}{2}\left(-\dfrac{3}{4}+\dfrac{5}{13}-\dfrac{9}{4}-\dfrac{8}{13}\right)=\dfrac{7}{2}\cdot\left(-3-\dfrac{3}{13}\right)=\dfrac{7}{2}\cdot\dfrac{-42}{13}=\dfrac{-147}{13}\)
b: \(=-12+\dfrac{8}{9}-\dfrac{5}{18}=\dfrac{-216}{18}+\dfrac{16}{18}-\dfrac{5}{18}=\dfrac{-205}{18}\)
c: \(=\dfrac{45}{4}-\dfrac{19}{7}-\dfrac{21}{4}=6-\dfrac{19}{7}=\dfrac{23}{7}\)
d: \(=\dfrac{-1}{4}\left(\dfrac{152}{11}+\dfrac{68}{11}\right)=\dfrac{-1}{4}\cdot20=-5\)
19 tháng 3 2020
\(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}-\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)
\(A=\frac{155-5\left(\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{403-13\left(\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}-\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{7}{91}+\frac{2}{10}-\frac{3}{10}}\)
\(A=\frac{155-5}{403-13}-\frac{3\left(\frac{1}{5}+\frac{1}{13}\right)-\frac{9}{10}}{\frac{7}{91}+\left(-\frac{1}{10}\right)}\)
\(A=\frac{5}{13}-\frac{\left(-\frac{9}{130}\right)}{\left(-\frac{3}{130}\right)}=\frac{5}{13}-\frac{\frac{9}{130}}{\frac{3}{130}}\)
\(A=\frac{5}{13}-\frac{9}{130}\cdot\frac{130}{3}\)
\(A=\frac{5}{13}-3=-\frac{34}{13}\)
\(B=\frac{30\cdot4^7\cdot3^{29}-5\cdot14^5\cdot2^{12}}{54\cdot6^{14}\cdot9^7-12\cdot8^5\cdot7^5}\)
\(B=\frac{30\cdot\left(2^2\right)^7\cdot3^{29}-5\cdot\left(2\cdot7\right)^5\cdot2^{12}}{54\cdot\left(2\cdot3\right)^{14}\cdot\left(3^2\right)^7-12\cdot\left(2^3\right)^5\cdot7^5}\)
\(B=\frac{30\cdot2^{14}\cdot3^{29}-5\cdot2^5\cdot7^5\cdot2^{12}}{54\cdot2^{14}\cdot3^{14}\cdot3^{14}-12\cdot2^{15}\cdot7^5}\)
\(B=\frac{30\cdot3^{29}-5\cdot2^{17}\cdot7^5}{54\cdot3^{28}-12\cdot2^{15}\cdot7^5}=\frac{30\cdot3-5\cdot2^2}{54-12}=\frac{5}{3}\)
T
10 tháng 9 2023
\(a,\dfrac{5^{16}\cdot27^7}{125^5\cdot9^{11}}=\dfrac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}\)
\(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)
\(b,\left(-0,2\right)^2\cdot5-\dfrac{2^{13}\cdot27^3}{4^6\cdot9^5}\)
\(=0,04\cdot5-\dfrac{2^{13}\cdot\left(3^3\right)^3}{\left(2^2\right)^6\cdot\left(3^2\right)^5}\)
\(=0,2-\dfrac{2^{13}\cdot3^9}{2^{12}\cdot3^{10}}\)
\(=0,2-\dfrac{2}{3}\)
\(=-\dfrac{7}{15}\)
\(c,\dfrac{5^6+2^2\cdot25^3+2^3\cdot125^2}{26\cdot5^6}\)
\(=\dfrac{5^6+2^2\cdot\left(5^2\right)^3+2^3\cdot\left(5^3\right)^2}{5^6\cdot26}\)
\(=\dfrac{5^6+4\cdot5^6+8\cdot5^6}{5^6\cdot26}\)
\(=\dfrac{5^6\left(1+4+8\right)}{5^6\cdot26}\)
\(=\dfrac{13}{26}\)
\(=\dfrac{1}{2}\)
#\(Toru\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
10 tháng 9 2023
\(a,\dfrac{5^{16}.27^7}{125^5.9^{11}}=\dfrac{\left(5^2\right)^8.9^7.3^7}{25^5.5^5.9^{11}}\\ =\dfrac{25^8.9^7.\left(3^2\right)^3.3}{25^5.\left(5^2\right)^2.5.9^{11}}=\dfrac{25^8.9^7.9^3.3}{25^5.25^2.5.9^{11}}\\ =\dfrac{25^8.9^{10}.3}{25^7.5.9^{11}}=\dfrac{25^7.9^{10}.25.3}{25^7.9^{10}.5.9}\\ =\dfrac{25.3}{5.9}=\dfrac{5.5.3}{5.3.3}=\dfrac{5}{3}\)
`a)3/5+13/5-5/9-3/7`
`=16/5-5/9-3/7`
`=698/315`
3/5 + 13 /5 -5/9 - 3/7 = 698/315