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Ta có :
\(2^{100}=\left(2^4\right)^{25}=16^{25}\)
\(3^{75}=\left(3^3\right)^{25}=27^{25}\)
\(5^{50}=\left(5^2\right)^{25}=25^{25}\)
Do \(16^{25}< 25^{25}< 27^{25}\)
\(\Rightarrow2^{100}< 5^{50}< 3^{75}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có :
75 ^2005 - 75^ 2004 và 75^2004-75
752005 -752004=1 ; 752004 -75=751929
Vì 1<751929 nên 7^2005-7^2004 <75^2004-75
![](https://rs.olm.vn/images/avt/0.png?1311)
a)Ta có : \(\hept{\begin{cases}3^{100}=\left(3^2\right)^{50}=9^{50}\left(1\right)\\2^{150}=\left(2^3\right)^{50}=8^{50}\left(2\right)\end{cases}}\)
Mà 9 > 8 => 950 > 850 => 3100 > 2150
Vậy 3100 > 2150
b) Ta có : \(\hept{\begin{cases}27^5=\left(3^3\right)^5=3^{15}\left(3\right)\\243^3=\left(3^5\right)^3=3^{15}\left(4\right)\end{cases}}\)
Từ (3) và (4) => 315 = 315 hay 275 = 2433
Vậy 275 = 2433 ( nên sửa lại 245 --> 243 nhá)
c) Ta có : \(81^{75}=\left(3^4\right)^{75}=3^{300}=\left(3^3\right)^{100}=27^{100}\)
Mà 27 < 30 => 27100 < 30100 hay 8175 < 30100
Vậy 8175 < 30100
a.
\(3^{100}=\left(3^2\right)^{50}=9^{50}\)
\(2^{150}=\left(2^3\right)^{50}=8^{50}\)
\(9^{50}>8^{50}\)
\(\Rightarrow3^{100}>2^{150}\)
b.
\(27^5=\left(3^3\right)^5=3^{15}\)
\(243^3=\left(3^5\right)^3=3^{15}\)
\(3^{15}=3^{15}\)
\(\Rightarrow27^5=243^3\)
c.
\(81^{75}=\left(3^4\right)^{75}=3^{300}=\left(3^3\right)^{100}=27^{100}\)
\(27^{100}< 30^{100}\Rightarrow81^{75}< 30^{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a=2^{100}=\left(2^4\right)^{25}=16^{25}\)
\(b=3^{75}=\left(3^3\right)^{25}=27^{25}\)
\(c=5^{50}=\left(5^2\right)^{25}=25^{25}\)
Vì: \(16^{25}< 25^{25}< 27^{25}\Rightarrow2^{100}< 5^{50}< 3^{75}\Rightarrow a< c< b\)
tíc mình nha
so sánh các số a,b,c
a=2100
b=375
c=550
\(=2^{100}=\left(2^{20}\right)^5\)
\(3^{75}=\left(3^{15}\right)^5\)
\(5^{50}=\left(5^{10}\right)^5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
275=(25)15=3215
530=(52)15=2515
Vì 3215>2515 nên 275>530
275=(25)15=3215
530=(52)15=2515
Vì 32>25 ,suy ra 3215>2515
Vậy 275>530
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn để ý: 81 = 3^4 (34) 27 = 3^3
64 = 2^6 256 = 2^8
Vậy \(\left(\frac{27}{64}\right)^{15}=\left(\frac{3^2}{8^2}\right)^{15}=\left(\frac{3}{8}\right)^{30};\left(\frac{81}{256}\right)^{10}=\left(\frac{3^4}{4^4}\right)^{10}=\left(\frac{3}{4}\right)^{40}\)
Vì 3/8 <3/4 ; 30<40 nên \(\left(\frac{3}{8}\right)^{30}<\left(\frac{3}{4}\right)^{40}\)
Hay (27/64)^15 < (81/256)^10
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ ta co \(50^{20}=\left(50^2\right)^{10}\)
\(\left(50^2\right)^{10}=2500^{10}< 2550^{10}\)
Hay \(50^{20}< 2550^{10}\)
b/ ta có \(3^{75}=\left(3^3\right)^{25}\)
\(5^{50}=\left(5^2\right)^{25}\)
\(\Rightarrow\left(3^3\right)^{25}=27^{25}\)
\(\Rightarrow\left(5^2\right)^{25}=25^{25}\)
Vay \(3^{75}>5^{50}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(8^4.16^5=\left(2^3\right)^4.\left(2^4\right)^5=2^{12}.2^{20}=2^{12+20}=2^{32}.\)
\(27^4.81^{10}=\left(3^3\right)^4.\left(3^4\right)^{10}=3^{12}.3^{40}=3^{52}.\)
*)ta thấy 8<3 và 30 < 20 => \(8^{30}< 3^{20}\)
84.165=(23)4.(24)5=212.220=212+20=232.
274.8110=(33)4.(34)10=312.340=352.
8<3 và 30 < 20 =>
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng a /b > 1 => a/b > a+m/b+m (a;b;m thuộc N*)
Ta có:
\(\frac{100^{10}-1}{100^{10}-3}>\frac{100^{100}-1+2}{100^{10}-3+2}\)
\(>\frac{100^{100}+1}{100^{10}-1}\)
\(81^{75}\)và \(30^{100}\)
Ta có:
\(\Leftrightarrow\)\(81^{75}=\left(3^4\right)^{75}=3^{300}=3^{100}.3^{200}\)
\(\Leftrightarrow\)\(30^{100}=\left(3.10\right)^{100}=3^{100}.10^{100}\)
So sánh \(3^{200}\)và \(10^{100}\)ta có:
\(\Leftrightarrow\)\(3^{200}=\left(3^2\right)^{100}=9^{100}< 10^{100}\)
Vì: \(3^{100}.9^{100}< 3^{100}.10^{100}\)
Nên: \(81^{75}< 30^{100}\)