a) 16¹⁹ và 8²⁵ 

Ta có :

16¹⁹ = ( 2⁴ )¹⁹ = 2⁷⁶

8²⁵ = ( 2³ )²⁵ = 2⁷⁵

Vì 2⁷⁶ > 2⁷⁵ Nên 16¹⁹ > 8²⁵

b) 5³⁶ và 11²⁴

Ta có :

5³⁶ = ( 5³ )¹² = 125¹²

11²⁴ = ( 11² )¹² = 121¹²

Vì 125¹² > 121¹² Nên 5³⁶ > 11²⁴

1.

\(M=3^0+3^1+......+3^{50}.\)

\(\Rightarrow3M=3+3^2+.......+3^{51}\)

\(\Rightarrow3M-M=\left(3+3^2+.......+3^{51}\right)-\left(3^0+3+.....+3^{50}\right)\)

\(\Rightarrow2M=3^{51}-1\)

\(\Rightarrow M=\frac{3^{51}-1}{2}\)

2.

\(a,\)Ta có : \(16^{19}=\left(2^4\right)^{19}=2^{76}\)

                     \(8^{25}=\left(2^3\right)^5=2^{75}\)

Vì \(2^{76}>2^{75}\Rightarrow16^{19}>8^{25}\)

\(b,\)Ta có : \(5^{36}=\left(5^3\right)^{12}=125^{12}\)

                      \(11^{24}=\left(11^2\right)^{12}=121^{12}\)

Vì \(125^{12}>121^{12}\Rightarrow5^{36}>11^{24}\)

a) 2³⁰ = (2³)¹⁰ = 8¹⁰

3²⁰ = (3²)¹⁰ = 9¹⁰

Vì 8¹⁰ < 9¹⁰ nên 2³⁰ < 3²⁰

b) 10²⁰ = (10²)¹⁰ = 100¹⁰

Vì 100¹⁰ > 90¹⁰ nên 10²⁰ > 90¹⁰

c) 25¹⁶ = (5²)¹⁶ = 5³²

125¹¹ = (5³)¹¹ = 5³³

Vì 5³² < 5³³ nên 25¹⁶ < 125¹¹

d) 81¹⁶ = (3⁴)¹⁶ = 3⁶⁴

16³² = (4²)³² = 4⁶⁴

Vì 3⁶⁴ < 4⁶⁴ nên 81¹⁶ < 16³²

⁶

a, Ta có:

\(2^{30}=2^{3.10}=\left(2^3\right)^{10}=8^{10}\)

\(3^{20}=3^{2.10}=\left(3^2\right)^{10}=9^{10}\)

Vì \(8^{10}< 9^{10}\) nên \(2^{30}< 3^{20}\)

b, Ta có:

\(10^{20}=10^{2.10}=\left(10^2\right)^{10}=100^{10}\)

Giữ nguyên \(90^{10}\)

Vì \(100^{10}>90^{10}\) nên \(10^{20}>90^{10}\)

a) \(9=6+3=6+\sqrt{9}\)

\(6+2\sqrt{2}=6+\sqrt{8}\)

\(\sqrt{8}< \sqrt{9}\) nên \(6+\sqrt{8}=6+2\sqrt{2}< 6+\sqrt{9}=9\)

b) \(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}=5+\sqrt{24}\)

\(3^2=9=5+4=5+\sqrt{16}\)

\(\sqrt{16}< \sqrt{24}\Rightarrow3^2< \left(\sqrt{2}+\sqrt{3}\right)^2\Rightarrow3< \sqrt{2}+\sqrt{3}\)

c) \(9+4\sqrt{5}=\left(2+\sqrt{5}\right)^2\)

\(16=\left(2+2\right)^2=\left(2+\sqrt{4}\right)^2\)

\(\sqrt{4}< \sqrt{5}\Rightarrow2+\sqrt{4}< 2+\sqrt{5}\Rightarrow\left(2+\sqrt{4}\right)^2=16< \left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)

d) \(\left(\sqrt{11}-\sqrt{3}\right)^2=14-2\sqrt{33}=14-\sqrt{132}\)

\(2^2=14-10=14-\sqrt{100}\)

\(\sqrt{100}< \sqrt{132}\Leftrightarrow-\sqrt{100}>-\sqrt{132}\Leftrightarrow14-\sqrt{100}>14-\sqrt{132}\)

\(\Rightarrow2>\sqrt{11}-\sqrt{3}\)

a) \(32^{50}\)và \(27^{51}\)

\(32^{50}=\left(32^2\right)^{25}=1024^{25}\)

\(27^{51}=\left(27^2\right)^{25}.27=729^{25}.27\)

Vì \(1024>729\)nên \(1024^{25}>729^{25}.27\)hay \(32^{50}>27^{51}\)

b) \(31^9\)và \(9^{16}\)

\(31^9=\left(91^3\right)^2=273^2\)

\(9^{16}=\left(9^2\right)^4=81^4=\left(81^2\right)^2=6561^2\)

Vì \(6561>273\)nên \(273^2< 6561^2\)hay \(31^9< 9^{16}\).

\(a,2^{150}=\left(2^3\right)^{50}=8^{50}< 9^{50}=\left(3^2\right)^{50}=3^{100}\\ b,2^{24}=\left(2^3\right)^8=8^8< 9^8=\left(3^2\right)^8=3^{16}\)

Nó hơi dài cậu chờ tí nka !

Mình ghi nhầm đề bài 1 tí đề bài là :

So sánh 2 số A và B biết : 

A = (3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1) và B = 3^32 - 1

A = (2-1)(2+1)(2^2 + 1 ) (2^4 + 1 ) ( 2^8 + 1) ( 2^16 + 1)

A = (2^2 - 1)(2^2 + 1 ) ( 2^4 + 1 )(2^8 + 1 )(2^16 + 1) 

A= ( 2^4 - 1 )( 2^4 + 1 )(2^8 + 1 )(2^16 + 1 )

A = (2^8 - 1 )(2^8 + 1 )(2^16 + 1 )

A = (2^16 - 1 )(2^16 + 1 )

A =  2^32 - 1 < 2^32 = B
Vậy A = B
k mik nka !