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17 tháng 8 2015
1) \(7.4^x=7.4^3\Leftrightarrow4^x=4^3;x=3\)
2) \(\frac{3}{2.5^x}=\frac{3}{2.5^{12}}\Leftrightarrow5^x=5^{12};x=12\)
\(2^x=2.2^8=2^9;x=9\)
4) \(5.3^x=7.3^5-2.3^5\Leftrightarrow5.3^x=3^5.\left(7-2\right)\)
\(\Leftrightarrow3^5.x=3^5.5;x=5\)
3 tháng 1 2022
\(\dfrac{2^{15}.9^3}{6^7.4^4}\\ =\dfrac{2^{15}.3^9}{3^7.2^7.2^8}\\ =\dfrac{2^{15}.3^9}{3^7.2^{15}}\\ =\dfrac{3^9}{3^7}\\ =3^2\\ =9\)
NT
2
DC
0
AM
4
28 tháng 12 2017
\(\dfrac{2^{15}.9^3}{6^7.4^4}=\dfrac{2^{15}.\left(3^2\right)^3}{3^7.2^7.2^8}=\dfrac{2^{15}.3^6}{3^7.2^{15}}=\dfrac{3^6}{3^7}=\dfrac{1}{3}\)
28 tháng 12 2017
\(\dfrac{2^{15}.9^3}{6^7.4^4}=\dfrac{2^{15}.3^6}{2^{15}.3^7}=\dfrac{1}{3}\)
8 tháng 10 2023
`#3107.101107`
A.
`12 + (-2).8`
`= 12 + (-16)`
`= 12 - 16`
`= -4`
Vì `-4 < 23 =>` đáp án A không thỏa mãn dk
B.
`8 - 4 + 37`
`= 4 + 37`
`= 41`
Vì `41 > 23 =>` đáp án B không thỏa mãn dk
C.
`7.4 + (-3)`
`= 28 + (-3)`
`= 28 - 3`
`= 25`
Vì `25 > 23 =>` đáp án C không thỏa mãn
D.
`9.8 - 72`
`= 72 - 72`
`= 0`
Vì `0 < 23 =>` đáp án D không thỏa mãn.
Vậy, không có đáp án nào thỏa mãn dk.
TA
4 tháng 9 2020
a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)
b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)
c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)
d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)
4 tháng 9 2020
Bài giải
a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)
b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)
c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)
d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)
A = \(\dfrac{3}{99.96}\) - \(\dfrac{3}{96.93}\) - \(\dfrac{3}{93.90}\) - ... - \(\dfrac{3}{7.4}\) - \(\dfrac{3}{4.1}\)
A = - (\(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + ...+ \(\dfrac{3}{90.93}\) + \(\dfrac{3}{92.96}\)) + \(\dfrac{3}{96.99}\)
A = - (\(\dfrac{1}{1}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + ..+ \(\dfrac{1}{90}\) - \(\dfrac{1}{93}\) + \(\dfrac{1}{93}\) - \(\dfrac{1}{96}\)) + \(\dfrac{1}{96}\) - \(\dfrac{1}{99}\)
A = - (\(\dfrac{1}{1}\) - \(\dfrac{1}{96}\)) + \(\dfrac{1}{96}\) - \(\dfrac{1}{99}\)
A = - 1 + \(\dfrac{1}{96}\) + \(\dfrac{1}{96}\)- \(\dfrac{1}{99}\)
A = - \(\dfrac{95}{96}\) + \(\dfrac{1}{96}\)- \(\dfrac{1}{99}\)
A = - \(\dfrac{47}{48}\) - \(\dfrac{1}{99}\)
A = - \(\dfrac{1567}{1584}\)