Câu a:

3^500 = (3^5)^100 = 243^100

7^300 = (7^3)^100 = 343^100 > 243^100

7^300 > 3^500

Câu b:

333^444 = (333^4)^111

444^333 = (444^3)^111

333^4 = 3^4.111^4 = 81.111^4

444^3 = 4^3.111^3 = 64.111^3 < 81.111^4

444^3< 333^4

(333^4)^111 > (444^3)^111

333^444 > 444^333

Ko biết

có \(777^{333}=\left(7.111\right)^{333}=7^{333}.111^{333}=7^{3.111}.111^{333}=\left(7^3\right)^{111}.111^{333}=343^{111}.111^{333}\)

mà \(333^{777}=\left(3.111\right)^{777}=3^{777}.111^{777}=\left(3^7\right)^{111}.111^{777}=2187^{111}.111^{777}\)

ta thấy \(343^{111}< 2187^{111},111^{333}< 111^{777}\)

=> \(343^{111}.111^{333}< 2187^{111}.111^{777}\)=> \(333^{777}< 777^{333}\)

vậy...

\(777^{333}=7^{333}.111^{333}=\left(7^3\right)^{111}.111^{333}=343^{111}.111^{333}\)

\(333^{777}=3^{777}.111^{777}=\left(3^7\right)^{111}.111^{777}=2187^{111}.111^{777}\)

Vì \(343^{111}< 2187^{111};111^{333}< 111^{777}\Rightarrow777^{333}< 333^{777}\)

Ta có: \(777^{333}=\left(777^3\right)^{111}=\left[\left(7.111\right)^3\right]^{111}=\left[7^3.111^3\right]^{111}\)

\(=\left[343.111^3\right]^{111}\)

\(333^{777}=\left(333^7\right)^{111}=\left[\left(3.111\right)^7\right]^{111}=\left[3^7.111^7\right]^{111}=\left(2187.111^7\right)^{111}\)

Vì \(343.111^3< 2187.111^7\Rightarrow777^3< 333^7\)

a , \(2^{3000}=\left(2^3\right)^{1000}=8^{1000}\)

      \(3^{2000}=\left(3^2\right)^{1000}=9^{1000}\)

Mà \(8^{1000}< 9^{1000}\)

\(\Rightarrow2^{3000}< 3^{2000}\)

b , \(777^{333}=\left(777^3\right)^{111}=\text{469097433}^{111}\)

     \(333^{777}=\left(333^7\right)^{111}=\text{36926037}^{111}\)

Mà 469097433¹¹¹>36926037¹¹¹

=> 777³³³>333⁷⁷⁷

2³⁰⁰⁰=(2³)¹⁰⁰⁰ =8¹⁰⁰⁰và 3²⁰⁰⁰=(3²)¹⁰⁰⁰ =9¹⁰⁰⁰

Vì 8 < 9 nên 8¹⁰⁰⁰ < 9¹⁰⁰⁰

do đó 2³⁰⁰⁰ <3²⁰⁰⁰
777³³³=(777³)¹¹¹
333⁷⁷⁷=(333⁷)¹¹¹
Vì 777³<333⁷ nên 777³³³<333⁷⁷⁷
Chắc chắn đúng đấy . Nhấn nút cảm ơn đi nhé

2⁵>3²

k mik nha

Học tốt

^_^

 giải rõ hơn đc ko bạn mà ko tính giá trị cụ thể

S6=15+17+19+21+...+151+153+155S6=15+17+19+21+...+151+153+155

Số các số hạng là:

(155−15):2+1=71(155−15):2+1=71

Vậy S6=(155+15).712=6035S6=(155+15).712=6035

S7=15+25+35+...+115S7=15+25+35+...+115

Số các số hạng là:

(115−15):10+1=11(115−15):10+1=11

Vậy S7=(115+15).112=715S7=(115+15).112=715

S4=24+25+26+...+125+126S4=24+25+26+...+125+126

Số các số hạng là:

(126−24):1+1=103

a, 444³³³=111³³³.4³³³=111³³³.64¹¹¹ 

333⁴⁴⁴=111⁴⁴⁴.3⁴⁴⁴=111⁴⁴⁴.81¹¹¹

suy ra 444³³³<333⁴⁴⁴

b,1²+2²+...+100²=1(2-1)+2(3-1)+...+100(101-1)

=(1.2+2.3+...+100.101)-(1+2+3...+100)

=A-5050

với A=1.2+2.3+...+100.101

3A=1.2.3+2.3.(4-1)+...+100.101.(102-99)

3A=1.2.3+2.3.4+...+100.101.102-(1.2.3+2.3.4+...+99.100.101)

=100.101.102

SUY RA  A=100.101.102/3=343400

thay vào ta có:

1²+2²+...+100²=A-5050=343400-5050=338350

Câu 1:

\(2^{1000}=\left(2^{10}\right)^{100}=1024^{100}>625^{100}=\left(5^4\right)^{100}=5^{400}\)

Câu 2:

a) \(2^n.8=512\)

\(\Leftrightarrow2^n.2^3=2^9\)

\(\Leftrightarrow2^n=2^6\Rightarrow n=6\)

b) \(\left(2n+1\right)^3=729\)

\(\Leftrightarrow\left(2n+1\right)^3=9^3\)

\(\Rightarrow2n+1=9\)

\(\Leftrightarrow2n=8\)

\(\Rightarrow n=4\)

Hè vẫn phải học chứ bạn :>

Câu 1.

2¹⁰⁰⁰ = ( 2⁵ )²⁰⁰ = 32²⁰⁰

5⁴⁰⁰ = ( 5² )²⁰⁰ = 25²⁰⁰

32 > 25 => 32²⁰⁰ > 25²⁰⁰ hay 2¹⁰⁰⁰ > 5⁴⁰⁰

Câu 2.

a) 2ⁿ . 8 = 512

<=> 2ⁿ . 2³ = 2⁹

<=> 2^{n + 3 = 9}

<=> n + 3 = 9

<=> n = 6

b) ( 2n + 1 )³ = 729

<=> ( 2n + 1 )³ = 3⁶

<=> ( 2n + 1 )³ = ( 3² )³ = 9³

<=> 2n + 1 = 9

<=> 2n = 8

<=> n = 4