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1) \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\)
2) \(3^{21}=3^{20}\cdot3=9^{10}\cdot3\)
\(2^{31}=2^{30}\cdot2=8^{10}\cdot2\)
mà \(9^{10}\cdot3>8^{10}\cdot2\)=> tự viết tiếp
3) đợi chút
430 = (43)10 = 6410 > 4810 = ( 2 . 24 )10 = ( 210 ) . ( 2410 ) > 3 . 2410
=> 230 + 330 + 430 > 3 . 2410
.
a.\(2^{225}=\left(2^3\right)^{75}=8^{75}\left(1\right)\)
\(3^{150}=\left(3^2\right)^{75}=9^{75}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow2^{225}< 3^{150}\)
b.\(2^{91}=2^{7.13}=\left(2^{13}\right)^7=8192^7\left(1\right)\)
\(5^{35}=5^{7.5}=\left(5^5\right)^7=3125^7\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow5^{35}< 2^{91}\)
c.\(9999^{10}=\left(99.101\right)^{10}=99^{10}.101^{10}>99^{10}.99^{10}=99^{20}\)
\(\Rightarrow9999^{10}>99^{20}\)
Bài 2:
\(b.\)\(75^{20}=\left(3.5^2\right)^{20}=\left(3^{20}.5^{10}\right).5^{30}=\left[̣\left(3^2\right)^{10}.5^{10}\right].5^{30}=45^{10}.5^{30}\)
\(\Rightarrowđpcm\)
Bài 3:
a,\(25^4.2^8=\left(5^2\right)^4.2^8=5^8.2^8=10^8\)
b. \(\frac{27^2}{25^3}=\frac{\left(3^3\right)^2}{\left(5^2\right)^3}=\frac{3^6}{5^6}=\left(\frac{3}{5}\right)^3\)\(:v\)
\(27^2.25^3=3^6.5^6=15^6\)( đề phòng you viết sai đề )
mãi ko thấy ai làm tớ làm giúp nì =)
\(\text{ta có:}\hept{\begin{cases}\frac{2002}{2003}< 1\\\frac{2005}{2004}>1\end{cases}}\Rightarrow\frac{2005}{2004}>\frac{2002}{2003}\Rightarrow-\frac{2005}{2004}< -\frac{2002}{2003}\)
\(\text{ta có: }\hept{\begin{cases}-\frac{1}{10^5}< 0\\\frac{-9}{-10}>0\end{cases}}\Rightarrow\frac{-1}{10^5}< \frac{-9}{-10}\)
\(A=1+3+3^2+3^3+...+3^{2016}\)
\(A=1+3\left(1+3^2+...+3^{2015}\right)\)
\(A=1+3\left(A-3^{2016}\right)\)
\(A=1+3A-3^{2017}\)
\(2A=3^{2017}-1\Rightarrow A=\frac{3^{2017}-1}{2}\)
\(A< B\)
Bài 2:
a: \(9^{20}=81^{10}\)
mà 81<9999
nên \(9^{20}< 9999^{10}\)
b: \(9^{20}=3^{40}\)
\(27^{13}=3^{39}\)
mà 40>39
nên \(9^{20}>27^{13}\)
a) 215.9466.83 =215.31236.26.29 =215.(32)436.315 =215.36.3236.315 =32=9
b) 310.11+310.539.24 =310(11+5)39.24 =39.3.2439.24 =3