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LN
1 tháng 11 2018
\(=\dfrac{5^{102}.\left(3^2\right)^{1009}}{3^{2018}.\left(5^2\right)^{50}}\)
\(=\dfrac{5^{102}.3^{2018}}{3^{2018}.5^{100}}=5^2=25\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 8 2023
\(\dfrac{5^{102}\cdot9^{1009}}{3^{2018}\cdot25^{50}}\)
\(=\dfrac{5^{102}\cdot\left(3^2\right)^{1009}}{3^{2018}\cdot\left(5^2\right)^{50}}\)
\(=\dfrac{5^{102}\cdot3^{2018}}{3^{2018}\cdot5^{100}}\)
\(=\dfrac{5^2\cdot1}{1\cdot1}\)
\(=25\)
N
21 tháng 11 2018
\(\left(3x-7\right)^{2009}=\left(3x-7\right)^{2007}\)
\(\Leftrightarrow\left(3x-7\right)^{2009}-\left(3x-7\right)^{2007}=0\)
\(\left(3x-7\right)^{2007}.\left[\left(3x-7\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(3x-7\right)^{2007}=0\\\left(3x-7\right)^2=1\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\\left(3x-7\right)=\pm1\end{cases}}}\)
=> \(x=\frac{7}{3},x=2,x=\frac{8}{3}\)
Vậy ...
21 tháng 11 2018
2/\(\frac{5^{102}.9^{1009}}{3^{2018}.25^{50}}=\frac{5^{100+2}.3^{2.1009}}{3^{2018}.5^{2.50}}=\frac{5^{100}.5^2.3^{2018}}{3^{2018}.5^{100}}=5^2=25\)
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
\(a,\dfrac{5^{16}.27^7}{125^5.9^{11}}=\dfrac{\left(5^2\right)^8.9^7.3^7}{25^5.5^5.9^{11}}\\ =\dfrac{25^8.9^7.\left(3^2\right)^3.3}{25^5.\left(5^2\right)^2.5.9^{11}}=\dfrac{25^8.9^7.9^3.3}{25^5.25^2.5.9^{11}}\\ =\dfrac{25^8.9^{10}.3}{25^7.5.9^{11}}=\dfrac{25^7.9^{10}.25.3}{25^7.9^{10}.5.9}\\ =\dfrac{25.3}{5.9}=\dfrac{5.5.3}{5.3.3}=\dfrac{5}{3}\)
19 tháng 3 2021
\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}:\sqrt{\dfrac{25}{9}}=\dfrac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}:\dfrac{5}{3}\)
\(=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}.\dfrac{5}{3}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1-5\right)}.\dfrac{5}{3}=\dfrac{1-3}{1-5}.\dfrac{5}{3}=\dfrac{1}{2}.\dfrac{5}{3}=\dfrac{5}{6}\)
19 tháng 3 2021
\(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\div\sqrt{\dfrac{25}{9}}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\div\dfrac{5}{4}\)
=\(\dfrac{2^{10}\cdot3^8\left(1-2\cdot3\right)}{2^{10}\cdot3^8\left(1+5\right)}\div\dfrac{5}{4}\)
=\(\dfrac{1-6}{1+5}\cdot\dfrac{4}{5}\)
=\(-\dfrac{5}{6}\cdot\dfrac{4}{5}\)
=\(-\dfrac{2}{3}\)
H
17 tháng 8 2021
a) 9/18 - (-7/12) + 13/32
= 13/12 + 13/32
= 143/96
b) (5/-8) + 14/25 - 6/10
= (-13/200) - 6/10
= -133/200
Chúc bạn học tốt!! ^^
17 tháng 8 2021
a: \(\dfrac{9}{18}-\dfrac{-7}{12}+\dfrac{13}{32}\)
\(=\dfrac{1}{2}+\dfrac{7}{12}+\dfrac{13}{32}\)
\(=\dfrac{48}{96}+\dfrac{56}{96}+\dfrac{39}{96}\)
\(=\dfrac{143}{96}\)
b: \(\dfrac{-5}{8}+\dfrac{14}{25}-\dfrac{6}{10}\)
\(=\dfrac{-125}{200}+\dfrac{112}{200}-\dfrac{120}{200}\)
\(=\dfrac{-133}{200}\)
\(\dfrac{5^{102}.9^{1009}}{3^{2018}.25^{50}}\)
\(=\dfrac{5^{102}.\left(3^2\right)^{1009}}{3^{2018}.\left(5^2\right)^{50}}\)
\(=\dfrac{5.1}{1.1}=5\)
\(\dfrac{5^{102}.9^{1009}}{3^{2018}.25^{50}}\)=\(\dfrac{5^{102}.3^{2018}}{3^{2018}.5^{100}}\) =5\(^2\) =25