N^200<5^300

=>(n^2)^100<(5^3)^100

=>n^2<5^3=125

=>n^2={0;4;9;...;121}

Ma n lon nhat=>n=11

Tick đi

\(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{12}\cdot\left(2^{18}+2^8\right)}{2^{12}\cdot\left(1+2^{10}\right)}=\frac{2^{18}+2^8}{1+2^{10}}\)

\(1;\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{12}\left(2^{18}+2^8\right)}{2^{12}\left(1+2^{22}\right)}=\frac{2^{18}+2^8}{1+2^{22}}\)

\(2;n^{200}< 5^{300}\Rightarrow\left(n^2\right)^{100}< 125^{100}\)

Vì n lớn nhất

\(\Rightarrow n^2=121=11^2\)

\(\Rightarrow n=11\)

\(n^{150}=\left(n^2\right)^{75};5^{225}=\left(5^3\right)^{75}=125^{75}\)

\(n^{150}< 5^{225}\) hay \(\left(n^2\right)^{75}< 125^{75}\)

=> \(n^2< 125\)

Nên: Số nguyên lớn nhất thỏa mãn điều kiện trên là n=11

Ta có: n^150 < 5^225
<=> n^2^75< 5^3^75
<=> n^2 < 5^3= 125
<=> n^2 ≤ 121
<=>n ≤11
mà n lớn nhất nên n=11
Vậy n=11

Ta có : \(n^{150}< 5^{225}\)

\(\Rightarrow\left(n^2\right)^{75}< \left(5^3\right)^{75}\)

=> n² < 125

=> n² = {1;4;9;16;25;36;49;64;81;100}

=> n = {1;2;3;4;5;6;7;8;9;10}

Vì n là lớn nhất 

Nên n = 10

Vậy n = 10

ta có:

\(n^{150}=\left(n^6\right)^{25};5^{225}=\left(5^9\right)^{25}\)

\(\Rightarrow n^6< 5^9\)

\(max\left(n\right)=11\)

Ta có:  \(n^{200}<5^{300}\)=> \(n^{2\cdot100}<5^{3\cdot100}=>\left(n^2\right)^{100}<\left(5^3\right)^{100}\Leftrightarrow n^2<5^3\Leftrightarrow n^2<125\)\(\Rightarrow n^2\in\left\{0;1;4;9;16;25;36;49;64;81;100;121\right\}\)

mà n >0

\(=>n\in\left\{1;2;3;4;5;6;7;8;9;10;11\right\}\)

mà n là số nguyên dương lớn nhất

=> n = 11

Vậy n =11

\(n^{150}< 5^{225}\)

\(\Leftrightarrow\left(n^{30}\right)^5< \left(5^{45}\right)^5\)

\(\Leftrightarrow n^{30}< 5^{45}\)

\(\Leftrightarrow\left(n^{10}\right)^3< \left(5^{15}\right)^3\)

\(\Rightarrow n^{10}< 5^{15}\)

\(\Leftrightarrow\left(n^2\right)^5< \left(5^3\right)^5\)

\(\Rightarrow n^2< 125\)

\(MAX_n\Rightarrow MAX_{n^2}\)

\(\Rightarrow n^2=121\Rightarrow n=11\)

1)

a) \(\frac{2^{15}\cdot9^4}{6^6\cdot8^3}=\frac{2^{15}\cdot3^8}{2^{15}\cdot3^6}=3^2=9\)

b) \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,4\right)^5\cdot2^5}{\left(0,4\right)^5\cdot0,4}=\frac{32}{0,4}=80\)

bài 4 : c1 \(3^{4000}\)và \(9^{2000}\)

\(\Leftrightarrow9^{2000}\Leftrightarrow\left(3^2\right)^2^{000}\Leftrightarrow3^{4000}\)

vì \(3^{4000}=3^{4000}\Leftrightarrow3^{4000}=9^{2000}\)

c2 

ta có 

\(3^{4000}=\left(3^4\right)^{1000}=81^{1000}\)

\(9^{2000}=\left(9^2\right)^{1000}=81^{1000}\)

vì \(81^{1000}=81^{1000}\Leftrightarrow3^{4000}=9^{2000}\)

bài 5 

\(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)

\(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)

vì \(8^{111}< 9^{111}\Leftrightarrow2^{332}< 3^{223}\)

3) M = 2²⁰¹⁰ - (2²⁰⁰⁹ + 2²⁰⁰⁸ + ....  + 2¹ + 2⁰)

Đặt N = 2²⁰⁰⁹ + 2²⁰⁰⁸ + ....  + 2¹ + 2⁰

=> 2N = 2²⁰¹⁰ + 2²⁰⁰⁹ + .... + 2² + 2¹

=> 2N - N = (2²⁰¹⁰ + 2²⁰⁰⁹ + .... + 2² + 2¹) - (2²⁰⁰⁹ + 2²⁰⁰⁸ + ....  + 2¹ + 2⁰)

=> N = 2²⁰¹⁰ - 1

Khi đó M = 2²⁰¹⁰ - (2²⁰¹⁰ - 1) = 1

4) C1 Ta có 3⁴⁰⁰⁰ = (3⁴)¹⁰⁰⁰ = 81¹⁰⁰⁰ = (9²)¹⁰⁰⁰ = 9²⁰⁰⁰ 

3⁴⁰⁰⁰ = 9²⁰⁰⁰

C2 Ta có : 3⁴⁰⁰⁰ = (3⁴)¹⁰⁰⁰ = 81¹⁰⁰⁰ (1)

Lại có 9²⁰⁰⁰ = (9²)¹⁰⁰⁰ = 81¹⁰⁰⁰ (2)

Từ (1) (2) => 3⁴⁰⁰⁰ = 9²⁰⁰⁰

5 Ta có 2³³² < 2³³³ = (2³)¹¹¹ = 8¹¹¹ < 9¹¹¹ = (3²)¹¹¹ = 3²²² < 3²²³

=> 2³³² < 3²²³

2) Ta có n¹⁵⁰ < 5²²⁵

=> (n⁵)⁷⁵ < (5³)⁷⁵

=> n⁵ < 5³

=> n⁵ < 125

Vì n là số nguyên lớn nhất => n = 2