Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(333^{333}=3^{333}.111^{333}\)
\(555^{222}=5^{222}.111^2\)
\(3^{333}=27^{111}>5^{222}=25^{111}\) (1)
\(111^{333}>111^{222}\)(2)
Từ (1) và (2) \(\rightarrow333^{333}>555^{222}\)
a/ \(3^{150}=\left(3^2\right)^{75}=9^{75}\)
\(2^{225}=\left(2^3\right)^{75}=8^{75}\)
\(9^{75}>8^{75}\Rightarrow3^{150}>2^{225}\)
b/
\(20162016^{10}=\left(2016.10001\right)^{10}=2016^{10}10001^{10}\)
\(2016^{20}=2016^{10}.2016^{10}\)
\(10001^{10}>2016^{10}\Rightarrow2016^{10}.10001^{10}>2016^{10}.2016^{10}\Rightarrow20162016^{10}>2016^{20}\)
c/ \(\frac{222^{333}}{333^{222}}=\frac{\left(222^3\right)^{111}}{\left(333^2\right)^{111}}=\frac{\left(2^3.111^3\right)^{111}}{\left(3^2.111^2\right)^{111}}=\left(\frac{8.111}{9}\right)^{111}\)
\(\frac{888}{9}>1\Rightarrow\left(\frac{888}{9}\right)^{111}>1\Rightarrow222^{333}>333^{222}\)
a) Ta có: 3^150 = 3^2.75 = (3^2)^75 = 9^75
2^225 = 2^3.75 = (2^3)^75 = 8^75
Vì 9 > 8 nên 9^75 > 8^75
Vậy 3^150 > 2^225
b) Ta có: 2016^20 = 2016^10+10 = 2016^10 . 2016^10
20162016^10 = (10001 . 2016)^10 = 10001^10 . 2016^10
Vì 2016^10 < 10001^10 nên 2016^10 . 2016^10 < 10001^10 . 2016^10
Vậy 2016^20 < 20162016^10
a/ 2225= (23)75 = 875
3150 = (32) 75 = 975
Vì 875 < 975 nên 2225 < 3150
b/ 3222 = (32)111 = 9111
2333 = (23)111 = 8111
vì 9111 > 8111 nên 3222 > 2333
2333=(23)111=8111
3222=(32)111=9111
Vì: 8<9 nên: 8111<9111
vậy: 2333<3222
b, 9920=(992)10=980110
Mà: 9801<9999 nên:
9920<999910
aTa có:
2333=(23)111=8111
3222=(32)111=9111
Do 8111<9111
=>2333<3222
b,Ta có:
9920=(992)10=980110
Do 980110 <999910
=>9920<999910
\(Ta\) \(có\) : \(222\equiv1\left(mod13\right)\) nên \(222^{333}\equiv1\left(mod13\right)\)
\(và\) \(333^2\equiv-1\left(mod13\right)\) nên \(333^{222}\equiv-1\left(mod13\right)\)
\(cộng\) \(lại\) \(ta\) \(có\) : \(222^{333}+333^{222}\equiv0\left(mod13\right)\) \(đpcm\)
Ta có:
\(222^{333}+333^{222}=111^{333}.2^{333}+111^{222}.3^{222}\)
\(=111^{222}\left[\left(111.2^3\right)^{111}+\left(3^2\right)^{111}\right]\)
\(=111^{222}\left(888^{111}+9^{111}\right)\)
\(\Rightarrow888^{111}+9^{111}\)
\(=\left(888+9\right)\left(888^{110}-888^{109}.9+...-888.9^{109}+9^{110}\right)\)
\(=13.69.\left(888^{110}-888^{109}.9+...-9^{109}+9^{110}\right)\)
\(=13.69.Q\)
\(\Rightarrow222^{333}+333^{222}⋮13\) (Đpcm)
\(222^{333}+333^{222}\)
\(=\left(222^3\right)^{111}+\left(333^2\right)^{111}⋮\left(222^3+333^2\right)=11051937⋮13\)
=> đpcm
Hằng đẳng thức: an - 1 = (a-1).[a(n-1) + a(n-2) +...+ 1] = (a-1).p (với n nguyên dương)
an + 1 = (a+1).[a(n-1) - a(n-2) +..+ 1] = (a+1).q (với n nguyên dương lẻ)
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
222333 - 1 = (222 - 1).p = 13.17.p
333222 + 1 = (333²)111 + 1 = 110889111 + 1 = (110889 + 1).q = 13.8530.q
222333 + 333222 = 222333 - 1 + 333222 + 1 = 13(17.p + 8530.q) chia hết cho 13
K NHÉ
222^3=10941048>333222
222^3<222^333
=>222^333>3332222