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Ta có: \(\frac{9^3.8^7}{6^9.16^2}=\frac{\left(3^2\right)^3.\left(2^3\right)^7}{2^9.3^9.\left(2^4\right)^2}\)=\(\frac{3^6.2^{21}}{2^9.3^9.2^8}=\frac{3^6.2^{21}}{2^{17}.3^9}=\frac{16}{27}\)
\(\frac{^{9^3}.^{^{8^7}}}{^{ }6^9.^{ }16^9}=\frac{729.2097152}{10077696.256}=\frac{1528823808}{2579890176}=\frac{16}{27}\)
Theo đề bài :
72x + 72x+2 = 2450
=> 72x . (1 + 72) = 2450
=> 72x . (1 + 49) = 2450
=> 72x . 50 = 2450
=> 72x = 49
=> 2x = 2
=> x = 1
Cho mình !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
72x+72x+2=2450
72x x (1+72)=2450
72x x 50 =2450
72x=2450:50
72x=49=72
suy ra 2x=2
suy ra x=1
ta có 8^9>7^9>6^9>5^9>4^9>3^9>2^9>1^9
->8^9+7^9+6^9+5^9+4^9+3^9+2^9+1^9<8^9×8=8^10<9^10
k mk nha
\(\frac{4^5.9^4-2.6^9}{2^{10}.3^6.20}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^6.2^2.5}=\frac{2^{10}.\left(3^8-3^9\right)}{2^{12}.3^6.5}=\frac{3^8-3^9}{2^2.3^6.5}=\frac{3^7\left(3-3^2\right)}{2^2.3^6.5}=\frac{3.\left(-6\right)}{20}\)\(=\frac{-18}{20}=\frac{-9}{10}\)
\(\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^6\cdot20}\)
\(=\frac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{12}\cdot3^6\cdot5}\)
\(=\frac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{12}\cdot3^6\cdot5}\)
\(=\frac{3^2\cdot\left(-2\right)}{2^2\cdot5}\)
\(=\frac{-9}{10}\)
\(10^{26}\) và \(9^{10}\)
Có: \(10>9\)
\(26>10\)
\(\Rightarrow10^{26}>9^{10}\)
C2: \(10^{26}=10^{10}.10^{16}\)
Vì: \(10^{10}>9^{10}\)
\(\Rightarrow10^{10}.10^{16}>9^{10}\)
\(\Rightarrow10^{26}>9^{10}\)
C1 10 ^ 26 = 100 ^ 25 = (100^5)^5 = 10000000000 ^ 5 > 81 ^ 5 = 9 ^10 => 10 ^ 26 > 9 ^ 10
C2 10 ^ 26 > 10^10 > 9^ 10 => 10 ^ 26 > 9 ^ 10
\(A=\frac{2^{19}3^9+5.2^{18}3^9}{3^9.2^9.2^{10}+3^{10}4^{10}}=\frac{3^9\left(2^{19}+5.2^{18}\right)}{3^92^{19}+3^{10}2^{20}}=\frac{3^9\left(2^{19}+5.2^{18}\right)}{3^9\left(2^{19}+3.2^{20}\right)}=\frac{2^{19}+5.2^{18}}{2^{19}+3.2^{20}}=\frac{2^{18}\left(2+5\right)}{2^{18}\left(2+3.2^2\right)}\)
\(=\frac{2+5}{2+3.2^2}=\frac{7}{14}=\frac{1}{2}\)
=(3^2)^3.(2^3)^7/(2.3)^9.(2^4)^2
=3^6.2^21/2^9.3^9.2^8
=1.2^4/1.3^3.1
=16/27
16/27