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a) Ta có: \(2^{225}=2^{3.75}=\left(2^3\right)^{75}=8^{75}\)
\(3^{150}=3^{2.75}=\left(3^2\right)^{75}=9^{75}\)
\(\Rightarrow8^{75}< 9^{75}\)\(\Rightarrow2^{225}< 3^{150}\)
b) Ta có : \(2^{91}=2^{7.13}=\left(2^{13}\right)^7=8192^7\)
\(5^{35}=5^{5.7}=\left(5^5\right)^7=3125^7\)
\(\Rightarrow8192^7>3125^7\)\(\Rightarrow2^{91}>3^{35}\)
c) Ta có: \(99^{20}=99^{2.10}=\left(99^2\right)^{10}=\left(99.99\right)^{10}\)
\(9999^{10}=\left(99.101\right)^{10}\)
Vì 99<101 \(\Rightarrow\left(99.99\right)^{10}< \left(99.101\right)^{10}\)\(\Rightarrow99^{20}< 9999^{10}\)
Ta có : \(2^{225}=\left[2^3\right]^{75}=8^{75}\)
\(3^{150}=\left[3^2\right]^{75}=9^{75}\)
Vì 8 < 9 => \(8^{75}< 9^{75}\)nên \(2^{225}< 3^{150}\)
Ta có : \(2^{91}>2^{90}=\left[2^5\right]^{18}=32^{18}>25^{18}=\left[5^2\right]^{18}=5^{36}>5^{35}\)
=> \(2^{91}>5^{35}\)
Ta có : \(9999^{10}=\left[99\cdot101\right]^{10}=99^{10}\cdot101^{10}>99^{10}\cdot99^{10}=99^{20}\)
Do đó : \(99^{20}< 9999^{10}\)
Cách khác : \(9999^{10}>9900^{10}=\left[99\cdot100\right]^{10}>\left[99^2\right]^{100}=99^{20}\)
Vậy : \(99^{20}< 9999^{10}\).
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 )
a \(2^{225}=8^{75}< 9^{75}=3^{150}\)
b: \(2^{91}=\left(2^{13}\right)^7=8192^7>3125^7=5^{35}\)
a) Ta có :
\(2^{225}=\left(2^3\right)^{75}=8^{75}\)
\(3^{150}=\left(3^2\right)^{75}=9^{75}\)
Vì \(8^{75}< 9^{75}\Leftrightarrow2^{225}< 3^{150}\)
b)Ta có :
\(2^{91}=\left(2^{13}\right)^7=8193^7\)
\(5^{35}=\left(5^5\right)^7=3125^7\)
Vì \(8193^7>3125^7\Leftrightarrow2^{91}>5^{35}\)
c) \(99^{20}=\left(99^2\right)^{10}=9801^{10}\)
Vì \(9801^{10}>9999^{10}\Leftrightarrow99^{20}>9999^{10}\)
a) Ta có :
2225=(23)75=8752225=(23)75=875
3150=(32)75=9753150=(32)75=975
Vì 875<975⇔2225<3150875<975⇔2225<3150
b)Ta có :
291=(213)7=81937291=(213)7=81937
535=(55)7=31257535=(55)7=31257
Vì 81937>31257⇔291>53581937>31257⇔291>535
c) 9920=(992)10=9801109920=(992)10=980110
Vì 980110>999910⇔9920>999910980110>999910⇔9920>999910