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NT
2
H9
HT.Phong (9A5)
CTVHS
12 tháng 8 2023
\(\left(2^3\cdot9^4+9^3\cdot45\right):\left(9^2\cdot10-9^2\right)\)
\(=\left(2^3\cdot3^8+3^6\cdot5\cdot3^2\right):\left[9^2\cdot\left(10-1\right)\right]\)
\(=\left(2^3\cdot3^8+3^8\cdot5\right):\left(9^2\cdot9\right)\)
\(=\left[3^8\cdot\left(2^3+5\right)\right]:9^3\)
\(=3^8\cdot13:3^6\)
\(=3^2\cdot13\)
\(=117\)
12 tháng 8 2023
\(\left(2^3.9^4+9^3.45\right):\left(9^2.10-9^2\right)\)
\(=\left(2^3.3^8+3^6.3^2.5\right):\left(3^4.10-3^4\right)\)
\(=\left(2^3.3^8+3^8.5\right):\left(3^4.\left(10-1\right)\right)\)
\(=3^8\left(2^3+5\right):3^4.9=3^8\left(2^3+5\right):3^6\)
\(=3^2.13=9.13=117\)
24 tháng 3 2021
\(A,\dfrac{7}{11}+\dfrac{5}{6}-\dfrac{-4}{11}-\dfrac{1}{6}\)
\(= (\dfrac{7}{11}+\dfrac{-4}{11})-(\dfrac{5}{6}-\dfrac{1}{6})\)
\(=\dfrac{3}{11}-\dfrac{2}{3}\)
\(= \dfrac{9}{33}-\dfrac{22}{33}\)
\(= \dfrac{-13}{33}\)
\(B,\dfrac{5}{9}.\dfrac{7}{3}+\dfrac{5}{9}.\dfrac{9}{13}-\dfrac{5}{9}:\dfrac{13}{5}\)
\(=(\dfrac{5}{9}.\dfrac{5}{9}:\dfrac{5}{9}).\dfrac{7}{3}+\dfrac{9}{13}-\dfrac{13}{5}\)
\(=(\dfrac{5}{9}.\dfrac{7}{3})+(\dfrac{9}{13}-\dfrac{13}{5})\)
\(=(\dfrac{5}{9}.\dfrac{21}{9})+(\dfrac{45}{65}-\dfrac{169}{65})\)
\(=\dfrac{35}{27}+\dfrac{-124}{65}\)
\(=\dfrac{2275}{1755}+\dfrac{-3348}{1755}\)
\(=\dfrac{-1073}{1755}\)
31 tháng 3 2021
a) Ta có: \(\dfrac{21}{25}\cdot\dfrac{11}{9}\cdot\dfrac{5}{7}\)
\(=\dfrac{21\cdot11\cdot5}{25\cdot9\cdot7}\)
\(=\dfrac{3\cdot7\cdot11\cdot5}{5\cdot5\cdot3\cdot3\cdot7}\)
\(=\dfrac{11}{5\cdot3}=\dfrac{11}{15}\)
8 tháng 4 2022
a: \(\dfrac{-11}{18}+\dfrac{12}{29}+\dfrac{-7}{18}+\dfrac{2020}{2021}+\dfrac{17}{29}\)
=(-11/18-7/18)+(12/29+17/29)+2020/2021
=2020/2021
b: \(\dfrac{-2}{3}\cdot\dfrac{4}{9}+\dfrac{5}{9}\cdot\dfrac{-2}{3}+\dfrac{4}{3}\)
=-2/3+4/3=2/3
29 tháng 1 2018
| 0 | = 0
| - 9 | = 9
| 7 | = 7
512 ( 2 - 128 ) - 128 ( - 512 )
= 512 . 2 - 512 . 128 - 128 . ( - 512 )
= 512 . 2 - 0
= 1024
LX
1
PT
0
\(A=1+\dfrac{1}{1+2}+...+\dfrac{1}{1+2+...+8}\)
\(=\dfrac{2}{2}+\dfrac{1}{2\cdot\dfrac{3}{2}}+...+\dfrac{1}{8\cdot\dfrac{9}{2}}\)
\(=\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+...+\dfrac{2}{8\cdot9}\)
\(=2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{8\cdot9}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
\(=2\left(1-\dfrac{1}{9}\right)=2\cdot\dfrac{8}{9}=\dfrac{16}{9}\)