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Ta có: 333^444= 111^444 x 3^444
444^333 = 111^333 x 4^333
Tách: 3^444 = (3^4)^111 =81^111 <=>4^333 = (4^3)^111 = 64^111
Mà: {111^444 > 111^333 (1)
{81^111 > 64^111 hay: (3^4)^111 > (4^3)^111 (2)
Từ (1) và (2) ta có:333^444 > 444^333
1. \(3^{54}=\left(3^2\right)^{27}=9^{27}\)
\(2^{81}=\left(2^3\right)^{27}=8^{27}\)
\(\text{Vì }9>8\text{ nên }9^{27}>8^{27}\)
\(\text{Vậy }3^{54}>2^{81}.\)
2. \(333^{444}=\left(3.111\right)^{444}=3^{444}.111^{444}=\left(3^4\right)^{111}.111^{444}=81^{111}.111^{444}\)
\(444^{333}=\left(4.111\right)^{333}=4^{333}.111^{333}=\left(4^3\right)^{111}.111^{333}=64^{111}.111^{333}\)
\(\text{Vì }81^{111}.111^{444}>64^{111}.111^{333}\text{ nên }333^{444}>444^{333}.\)
3. \(10^{30}=\left(10^3\right)^{10}=1000^{10}\)
\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)
Vì 1000 < 1024 nên 100010 < 102410.
Vậy \(10^{30}<2^{100}.\)
Để 5n8 chia hết cho 3
Thì 5 + n + 8 chia hết cho 3
=> 13 + n chia hết cho 3
=> n = 2,5,8
Bài 1 :
a. " \(\in\){ 2 ; 5 ; 8 }
b. " \(\in\){ 0 ; 9 }
c. ' = 5
d. 0
Câu dưới nha :
Có : A = 3^450 = (3^3)^150 = 27^150
5^300 = (5^2)^150 = 25^150
Vì 27^150 > 25^150 => 3^450 > 5^300
k mk nha
theo đề, ta có: A= 333^444=(111.3)^4.111=(111^4.3^4)^111=(111^4.81)^111
B=444^333=(111.4)^111.3=(111^3.4^3)^111=(111^3.64)^111
Vì 111^4.81 >111^3.64 nên A>B
cho mình 1k nhé bạn
a)\(1024^9=\left(2^{10}\right)^9=2^{90}< 2^{100}\)
b)\(27^{11}=\left(3^3\right)^{11}=3^{33}>3^{32}=\left(3^4\right)^8=81^8\)
c)\(5^{20}=\left(5^2\right)^{10}=25^{10}\)
\(3^{30}=\left(3^3\right)^{10}=27^{10}\)
Ta có: \(25^{10}< 27^{10}\)
\(\Rightarrow5^{20}< 3^{30}\)
d) tương tự
e) \(78^{12}-78^{11}=78^{11}.\left(78-1\right)=78^{11}.77\)
\(78^{11}-78^{10}=78^{10}.\left(78-1\right)=78^{10}.77\)
Ta có: \(78^{11}.77>78^{10}.77\)
\(\Rightarrow78^{12}-78^{11}>78^{11}-78^{10}\)
f) \(333^{444}=\left[\left(111.3\right)^4\right]^{111}=\left(111^4.3^4\right)^{111}=\left(111^4.81\right)^{111}\)
\(444^{333}=\left[\left(111.4\right)^3\right]^{111}=\left(111^3.4^3\right)^{111}=\left(111^3.64\right)^{111}\)
Ta có: \(111^4.81>111^3.64\)
\(\Rightarrow\left(111^4.81\right)^{111}>\left(111^3.64\right)^{111}\)
\(\Rightarrow333^{444}>444^{333}\)
Tham khảo nhé~
a) Ta có :
\(1024^9=\left(2^{10}\right)^9=2^{90}\)
Vì \(2^{100}>2^{90}\)\(\Rightarrow\)\(2^{100}>1024^9\)
Vậy \(2^{100}>1024^9\)
b) Ta có :
\(27^{11}=\left(3^3\right)^{11}=3^{33}\)
\(81^8=\left(3^4\right)^8=3^{32}\)
Vì \(3^{33}>3^{32}\)\(\Rightarrow\)\(27^{11}>81^8\)
Vậy \(27^{11}>81^8\)
c) Ta có :
\(5^{20}=\left(5^2\right)^{10}=25^{10}\)
\(3^{30}=\left(3^3\right)^{10}=27^{10}\)
Vì \(25^{10}< 27^{10}\)\(\Rightarrow\)\(5^{20}< 3^{30}\)
Vậy \(5^{20}< 3^{30}\)
d) Ta có :
\(2^{300}=\left(2^3\right)^{100}=8^{100}\)
\(3^{200}=\left(3^2\right)^{100}=9^{100}\)
Vì \(8^{100}< 9^{100}\)\(\Rightarrow\)\(2^{300}< 3^{200}\)
Vậy \(2^{300}< 3^{200}\)
e) Ta có :
\(78^{12}-78^{11}=78^{11}.\left(78-1\right)=78^{11}.77\)
\(78^{11}-78^{10}=78^{10}\left(78-1\right)=78^{10}.77\)
Vì \(78^{11}>78^{10}\)\(\Rightarrow78^{11}.77>78^{10}.77\)
Hay \(78^{12}-78^{11}>78^{11}-78^{10}\)
Vậy \(78^{12}-78^{11}>78^{11}-78^{10}\)
f) Ta có :
\(333^{444}=\left(333^4\right)^{111}=\left[\left(3.111\right)^4\right]^{111}=\left(3^4.111^4\right)^{111}=\left(81.111^4\right)^{111}\)
\(444^{333}=\left(444^3\right)^{111}=\left[\left(4.111\right)^3\right]^{111}=\left(4^3.111^3\right)^{111}=\left(64.111^3\right)^{111}\)
Vì \(81.111^4>64.111^3\)\(\Rightarrow\)\(\left(81.111^4\right)^{111}>\left(64.111^3\right)^{111}\)
Hay \(333^{444}>444^{333}\)
Vậy \(333^{444}>444^{333}\)
_Chúc bạn học tốt_