a) \(2^3=8\) ; \(3^2=9\)

=> \(2^3< 3^2\)

b) \(3^{210}\cdot3^{10}=3^{210+10}=3^{220}>3^{215}\)

=> \(3^{215}< 3^{210}.3^{10}\)

a,\(2^3\)và \(3^2\)

\(2^3=8\)\(3^2=9\)

Vì \(8< 9\Rightarrow2^3< 3^2\)

Vậy....

b,\(3^{215}\)và \(3^{210}.3^{10}\)

\(3^{215}\)và \(3^{220}\)

\(3^{215}< 3^{220}\Rightarrow3^{215}< 3^{210}.3^{10}\)

Vậy...

a) 2³ = 8 ; 3² = 9

Vì 8 < 9 => 2³ < 3²

b) 3²¹⁰.3¹⁰ = 3²¹⁰⁺¹⁰ = 3²²⁰ > 3²¹⁵

=> 3²¹⁵ < 3²¹⁰.3¹⁰

4 ^ 280 > 5 ^ 210

\(4^{280}>5^{210}\)?

S = 2⁰ + 2 + 2² + 2³ + 2⁴ +2⁵ +2⁶ +2⁷ + 2⁸ + 2⁹
2S = 2.( 2⁰ + 2 + 2² + 2³ + 2⁴ +2⁵ +2⁶ +2⁷ + 2⁸ + 2⁹)

2S = 2 +  2² + 2³ + 2⁴ +2⁵ +2⁶ +2⁷ + 2⁸ + 2⁹ + 2¹⁰

S = (2 +  2² + 2³ + 2⁴ +2⁵ +2⁶ +2⁷ + 2⁸ + 2⁹ + 2¹⁰) - (2⁰ + 2 + 2² + 2³ + 2⁴ +2⁵ +2⁶ +2⁷ + 2⁸ + 2⁹)

S = 2¹⁰ - 2⁰

ta có: 5 x 2⁸ = ( 4 + 1) x 2⁸ = 4 . 2⁸ + 2⁸ = 2² . 2⁸ + 2⁸ = 2¹⁰ + 2⁸

vì 2¹⁰ - 2⁰ < 2¹⁰ + 2⁸ nên S < 5 x 2⁸

\(2A=2^2+2^4+2^5+...+2^{2011}\)

\(\Leftrightarrow A=2^{2011}-2< 2^{2022}-2\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}=\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1\)

2S=2(1+2+2²+...+2⁹)

2S=2+2²+...+2¹⁰

2S-S=(2+2²+...+2¹⁰)-(1+2+2²+...+2⁹)

S=2¹⁰-1=1024-1=1023

5*2⁸=5*256=1280.Vì 1280>1023

=>5*2⁸>2¹⁰-1 <=> 5*2⁸>S