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\(=\dfrac{5^{102}.\left(3^2\right)^{1009}}{3^{2018}.\left(5^2\right)^{50}}\)
\(=\dfrac{5^{102}.3^{2018}}{3^{2018}.5^{100}}=5^2=25\)
\(\left(3x-7\right)^{2009}=\left(3x-7\right)^{2007}\)
\(\Leftrightarrow\left(3x-7\right)^{2009}-\left(3x-7\right)^{2007}=0\)
\(\left(3x-7\right)^{2007}.\left[\left(3x-7\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(3x-7\right)^{2007}=0\\\left(3x-7\right)^2=1\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\\left(3x-7\right)=\pm1\end{cases}}}\)
=> \(x=\frac{7}{3},x=2,x=\frac{8}{3}\)
Vậy ...
2/\(\frac{5^{102}.9^{1009}}{3^{2018}.25^{50}}=\frac{5^{100+2}.3^{2.1009}}{3^{2018}.5^{2.50}}=\frac{5^{100}.5^2.3^{2018}}{3^{2018}.5^{100}}=5^2=25\)
Ta có : \(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)=\frac{2018}{2017}-2019.2-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
\(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)\)
\(=\frac{2018}{2017}-2018.\frac{2019}{1009}-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-2.2019-\frac{2019}{2017}+2.2019\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
(1/3)50. (-9)25 - 2/3 : 4
= (1/3)50 . [(-3)2]25 - 2/3 . 1/4
= (1/3)50.(-3)50- 1/6
= (1/3 . -3 )50 - 1/6
= (-1)50- 1/6
= 1 - 1/6
= 5/6
theo đề ta có
=\(\left(\frac{1}{3^{ }}\right)^{50}.\left(-9\right)^{25}-\frac{2}{3}.\frac{1}{4}\)
=\(\left(\frac{1}{3}\right)^{50}.\left(\frac{1}{3}\right)^{25}.\left(-27\right)^{25}-\frac{1}{6}\)
=\(\left(\frac{1}{3}\right)^{50+25}.\)
Lời giải :
a ) \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)
\(=2,5\)
b ) \(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)
\(=\dfrac{3}{7}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)\)
\(=\dfrac{3}{7}\left(19-33\right)\)
\(=\dfrac{3}{7}\left(-14\right)\)
\(=-6\)
c ) \(9\left(-\dfrac{1}{3}\right)^3+\dfrac{1}{3}\)
\(=9\left(-\dfrac{1}{27}\right)+\dfrac{1}{3}\)
\(=-\dfrac{1}{3}+\dfrac{1}{3}\)
\(=0\)
d ) \(15\dfrac{1}{4}\div\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}\div\left(-\dfrac{5}{7}\right)\)
\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right)\div\left(-\dfrac{5}{7}\right)\)
\(=-10\left(-\dfrac{7}{5}\right)\)
\(=14\)
\(\dfrac{5^{102}\cdot9^{1009}}{3^{2018}\cdot25^{50}}\)
\(=\dfrac{5^{102}\cdot\left(3^2\right)^{1009}}{3^{2018}\cdot\left(5^2\right)^{50}}\)
\(=\dfrac{5^{102}\cdot3^{2018}}{3^{2018}\cdot5^{100}}\)
\(=\dfrac{5^2\cdot1}{1\cdot1}\)
\(=25\)
\(\dfrac{5^{102}.9^{1009}}{3^{2018}.25^{50}}\)
\(=\dfrac{5^{102}.\left(3^2\right)^{1009}}{3^{2018}.\left(5^2\right)^{50}}\)
\(=\dfrac{5.1}{1.1}=5\)
\(\dfrac{5^{102}.9^{1009}}{3^{2018}.25^{50}}\)=\(\dfrac{5^{102}.3^{2018}}{3^{2018}.5^{100}}\) =5\(^2\) =25