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a)Ta có:
\(2^{24}=\left(2^6\right)^4=64^4\)
\(3^{16}=\left(3^4\right)^4=81^4\)
Vì 644<814 nên 224<316
b)Ta có:
\(2^{30}=\left(2^3\right)^{10}=8^{10}\)
\(3^{20}=\left(3^2\right)^{10}=9^{10}\)
Vì 810<910 nên 230<320
a, 224 = 23.8 = (23)8 = 88
316 = 32.8 = (32)8 = 98
Có 88 < 98
=> 224 < 316
b, 230 = 23.10 = (23)10 = 810
320 = 32.10 = (32)10 = 910
Vì 810 < 910
=> 230 < 320
\(2^{20}+3^{30}+4^{30}=4^{10}+9^{10}+64^{10}<64^{10}+64^{10}+64^{10}=3.64^{10}\)
\(324^{10}>320^{10}=\left(5.64\right)^{10}=5^{10}.64^{10}>3.64^{10}\)
\(\Rightarrow2^{20}+3^{30}+4^{30}<324^{10}\)
a) 220= 22.10= ( 22)10=410
330= 33.10=(33)10= 2710
Vì 410 < 2710
=> 220 < 330
b) 2505= 25.101= (25)101= 32101
5 202= 52.101= (52)101= 25101
Vì 32101>25101
=> 2505>5202
\(a,2^{20}=\left(2^2\right)^{10}=4^{10}\)(1)
\(3^{30}=\left(3^3\right)^{10}=27^{10}\)(2)
Từ (1) và (2)
\(\Rightarrow3^{30}>2^{20}\)
\(b,2^{505}=\left(2^5\right)^{101}=32^{101}\)(1)
\(5^{202}=\left(5^2\right)^{101}=25^{101}\)(2)
Từ(1) và (2)
\(2^{505}>5^{202}\)
\(2^0+2^1+2^2+2^3+...+2^{50}=1+2+2.2+2^2.2+...+2^{49}.2\)
\(=1+2\left(1+2+2^2+2^3+...+2^{49}\right)\)
\(=1+2\left(2^{50}-1\right)\)
\(=1+2^{51}-2\)
\(=2^{51}-1< 2^{51}\)
Vậy \(2^0+2^1+2^2+2^3+...+2^{50}< 2^{51}\)
Ý trc mình ko biết sorry bạn nhiều
T i c k cho mình nha mình mới có 4 điểm, thanks
a) Ta có: \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)
b) Ta có: \(2^{31}=\left(2\frac{31}{21}\right)^{21}=2,7822^{21}< 3^{21}\Rightarrow2^{31}< 3^{21}\)
c) Ta có: \(3^{30}=\left(3^3\right)^{10}=27^{10}\)
\(2^{30}=\left(2^3\right)^{10}=8^{10}\)
\(4^{30}=\left(4^3\right)^{10}=64^{10}\)
Lại có: \(3.24^{10}=2.24^{10}+24^{10}\Rightarrow24^{10}< 27^{10}\left(1\right)\)
\(2.24^{10}< 48^{10}< 64^{10}\left(2\right)\)
Từ 1,2 => \(24^{10}+2.24^{10}< 27^{10}+64^{10}\Rightarrow3.24^{10}< 8^{10}+27^{10}+64^{10}\)
\(\Rightarrow3.24^{10}< 3^{30}+2^{30}+4^{30}\)
ta có \(2^{30}=\left(2^3\right)^{10}=8^{10}\)
\(3^{30}=\left(3^3\right)^{10}=27^{10}\)
\(4^{30}=\left(4^3\right)^{10}=64^{10}\)
ta có \(3^{20}=\left(3^2\right)^{10}=9^{10}\)
\(6^{20}=\left(6^2\right)^{10}=36^{10}\)
\(8^{20}=\left(8^2\right)^{10}=64^{10}\)
\(\Rightarrow2^{30}+3^{30}+4^{30}=8^{10}+27^{10}+64^{10}\)
\(\Rightarrow3^{20}+6^{20}+8^{20}=9^{10}+36^{10}+64^{10}\)
Xét \(8^{10}<9^{10}\) (1)
\(27^{10}<36^{10}\)(2)
\(64^{10}=64^{10}\)(3)
từ (1)(2)(3)\(\Leftrightarrow8^{10}+27^{10}+64^{10}<9^{10}+36^{10}+64^{10}\)
\(\Rightarrow2^{30}+3^{30}+4^{30}<3^{20}+6^{20}+8^{20}\)
Ta có: \(2^{30}+3^{30}+4^{30}=\left(2^3\right)^{10}+\left(3^3\right)^{10}+\left(4^3\right)^{10}=8^{10}+27^{10}+64^{10}\)
\(3^{20}+6^{20}+8^{20}=\left(3^2\right)^{10}+\left(6^2\right)^{10}+\left(8^2\right)^{10}=9^{10}+36^{10}+64^{10}\)
Vì \(8< 9\)\(\Rightarrow8^{10}< 9^{10}\)
mà \(27< 36\)\(\Rightarrow27^{10}< 36^{10}\)
\(\Rightarrow8^{10}+27^{10}< 9^{10}+36^{10}\)
\(\Rightarrow8^{10}+27^{10}+64^{10}< 9^{10}+36^{10}+64^{10}\)
hay \(2^{30}+3^{30}+4^{30}< 3^{20}+6^{20}+8^{20}\)
so sánh: 2^30 + 3^30 + 4^30 và 3^20 + 6^20 + 8^20
2^30 = ( 2^3)^10 = 8^ 10
3^30 = (3^3)^10 = 27^10
4^30 = (4^3)^10 = 64^10
3^20 = (3^2)^10 = 9^10
6^20 = (6^2) = 36^10
8^20 = (8^2)^10 = 84^10
vì 9^10 > 8^10
36^10 > 27^10
84^10 > 64^10
=> 2^30 + 3^30 + 4^30 < 3^20 + 6^20 + 8^20