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H9
HT.Phong (9A5)
CTVHS
1 tháng 8 2023
\(\dfrac{5^{102}\cdot9^{1009}}{3^{2018}\cdot25^{50}}\)
\(=\dfrac{5^{102}\cdot\left(3^2\right)^{1009}}{3^{2018}\cdot\left(5^2\right)^{50}}\)
\(=\dfrac{5^{102}\cdot3^{2018}}{3^{2018}\cdot5^{100}}\)
\(=\dfrac{5^2\cdot1}{1\cdot1}\)
\(=25\)
OK
2
FC
20 tháng 11 2018
\(\dfrac{5^{102}.9^{1009}}{3^{2018}.25^{50}}\)
\(=\dfrac{5^{102}.\left(3^2\right)^{1009}}{3^{2018}.\left(5^2\right)^{50}}\)
\(=\dfrac{5.1}{1.1}=5\)
20 tháng 11 2018
\(\dfrac{5^{102}.9^{1009}}{3^{2018}.25^{50}}\)=\(\dfrac{5^{102}.3^{2018}}{3^{2018}.5^{100}}\) =5\(^2\) =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/\(\frac{5^{102}.9^{1009}}{3^{2018}.25^{50}}=\frac{5^{100+2}.3^{2.1009}}{3^{2018}.5^{2.50}}=\frac{5^{100}.5^2.3^{2018}}{3^{2018}.5^{100}}=5^2=25\)
29 tháng 6 2023
a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)
=1-2/4=1/2
b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)
c: x-y=0 nên x=y
\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)
=2019
DT
0
20 tháng 10 2023
o: \(\dfrac{\left(-1\right)^6\cdot3^5\cdot4^3}{9^2\cdot2^5}=\dfrac{3^5\cdot2^6}{2^5\cdot3^4}=\dfrac{3^5}{3^4}\cdot\dfrac{2^6}{2^5}=3\cdot2=6\)
s: \(\dfrac{\dfrac{2}{7}+\dfrac{2}{5}+\dfrac{2}{17}-\dfrac{2}{25}}{\dfrac{3}{14}+\dfrac{3}{10}+\dfrac{3}{34}-\dfrac{3}{50}}\)
\(=\dfrac{2\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{25}\right)}{\dfrac{3}{2}\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{25}\right)}\)
\(=2:\dfrac{3}{2}=\dfrac{4}{3}\)
t: \(\sqrt{\dfrac{4}{9}}-\dfrac{1}{2}:\left|-\dfrac{2}{3}\right|\)
\(=\dfrac{2}{3}-\dfrac{1}{2}:\dfrac{2}{3}\)
\(=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8-9}{12}=-\dfrac{1}{12}\)
LB
1
20 tháng 10 2019
\(\frac{5^{102}\cdot9^{1000}}{3^{2018}\cdot25^{50}}=\frac{5^{102}\cdot3^{2000}}{3^{2018}\cdot5^{100}}=\frac{5^2}{3^{18}}\)
VN
1
19 tháng 12 2018
\(\dfrac{2018^{2019}.4^{2018}}{1009^{2019}.8^{2019}}\)
=\(\dfrac{2018^{2019}.4^{2018}}{1009^{2019}.\left(2.4\right)^{2019}}\)
=\(\dfrac{2018^{2019}.4^{2018}}{1009^{2019}.2^{2019}.4^{2019}}\)
=\(\dfrac{2018^{2019}.4^{2018}}{\left(1009.2\right)^{2019}.4^{2019}}\)
=\(\dfrac{2018^{2019}.4^{2018}}{2018^{2019}.4^{2019}}\)
=\(\dfrac{1}{4}\)
\(=\dfrac{5^{102}.\left(3^2\right)^{1009}}{3^{2018}.\left(5^2\right)^{50}}\)
\(=\dfrac{5^{102}.3^{2018}}{3^{2018}.5^{100}}=5^2=25\)
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