(2222^1)^1111 va (1111^2)1111

(2 x 1111)^1 va (1 x 11111)^2

2 x 2222^1 va 1 x 1111^2

2222 x 1111^2  va  1111 x 2222^1

ket luan : 2222^1111 va 1111^2222

ta có 1111²²²²=(1111²)¹¹¹¹

=1234321¹¹¹¹

vì 2222¹¹¹¹<1234321¹¹¹¹

nên 2222¹¹¹¹<1111²²²²

Ta thấy : \(2222^{3333}vs2^{300}:\hept{\begin{cases}2222>2\\3333>300\end{cases}\Rightarrow2222^{3333}>2^{300}}\)

Ta thấy : \(2222^{1111}=1111^{1111}.2^{1111}< 1111^{1111}.1111^{1110}=1111^{2221}\)

Ta thấy : \(54^{10}=\left(3^3\right)^{10}.2^{10}=3^{30}.2^{10}=3^{12}.3^{18}.2^{10}>3^{12}.7^{12}=21^{12}.\)

\(2222^{1111}=2^{1111}.1111^{1111}\)

\(1111^{2222}=1111^{1111}.1111^{1111}\)

Vì: \(2^{1111}< 1111^{1111}\)

\(\Rightarrow2222^{1111}< 1111^{2222}\)

1111²²²² = (1111²)¹¹¹¹

mà 1111² > 2222

nên 1111²²²² > 2222¹¹¹¹

cần giải thích thì nói, mình chỉ cho

Có tận cùng là 6.

Đặt 11......1 (n chữ số 1 ) =a ( a thuộc N )

=> 2222.....2(n chữ số 2) =2a

100....0(n chữ số 0) = 9a+1

=> 1111....1(2n chữ số 1) = a.(9a+1)+a

Khi đó : A = a.(9a+1)+a-2a = 9a^2+a+a-2a=9a^2 = (3a)^2 là số chính phương)

=> ĐPCM

Mình không hiểu luôn ak !!!!@@@

\(\left(\frac{13}{5}x\frac{11}{20}-\frac{11}{10}\right):\frac{11}{40}=\left(\frac{143}{100}-\frac{110}{100}\right):\frac{11}{40}=\frac{33}{100}:\frac{33}{120}=\frac{33}{100}x\frac{120}{33}=\frac{6}{5}\)

thank gì cho đúng đi

 

lười thế

1

2⁵⁰⁰ và 3³⁰⁰

2⁵⁰⁰= (2⁵)¹⁰⁰= 32¹⁰⁰

3³⁰⁰= (3³)¹⁰⁰= 27¹⁰⁰

Vì 32¹⁰⁰> 27¹⁰⁰ nên 2⁵⁰⁰ > 3³⁰⁰

Vậy...

gom vô

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(1111+8889)+(2222+7778)+(4444+5556)

=10000+10000+10000
Đây nha