a) \(A=1+2+2^2+2^3+...+2^{100}\) \(B=2^{201}\)

\(2A=2\left(1+2+2^2+2^3+...+2^{100}\right)\)

\(2A=2+2^2+2^3+2^4+...+2^{201}\)

\(2A-A=\left(2+2^2+2^3+2^4+...+2^{201}\right)-\left(1+2+2^2+2^3+...+2^{100}\right)\)

\(2A-A=2^{101}-1\)

\(A=2^{201}-1\)

Ta có 2²⁰¹ > 2²⁰¹ - 1 => B > A => 2²⁰¹ > 1 + 2 + 2² + 2³ +...+ 1¹⁰⁰

b) 2¹⁰⁰ = 2³¹ . 2⁶³ . 2⁶ = 2³¹ . (2⁹)⁷ . (2²)³ = 2³¹ . 512⁷ . 4³ (1)

10³¹ = 2³¹ . 5²⁸ . 5³ = 2³¹ . (5⁴)⁷ . 5³ = 2³¹ . 625⁷ . 5³ (2)

Từ (1) , (2) => 2³¹ . 512⁷ . 4³ < 2³¹ . 625⁷ . 5³ ( vì 512⁷ < 625⁷ và 4³ < 5³ )

=> 2¹⁰⁰ < 10³¹

e) Ta có:

2¹⁰⁰ = (2¹⁰)¹⁰ = 1024¹⁰

10³⁰ = (10³)¹⁰ = 1000¹⁰
Vì 1024¹⁰ > 1000¹⁰ => 2¹⁰⁰ > 10³⁰

Ta có A=\(\frac{1}{5}\)+\(\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)\)+\(\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)\)

Ta lại có: \(\frac{1}{5}=\frac{1}{5}\)

\(\frac{1}{13}=\frac{1}{13},\frac{1}{13}>\frac{1}{14},\frac{1}{13}>\frac{1}{15}\)

\(\frac{1}{61}=\frac{1}{61},\frac{1}{61}>\frac{1}{62},\frac{1}{61}>\frac{1}{63}\)

\(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\)<\(\frac{1}{5}+\frac{1}{13}+\frac{1}{13}+\frac{1}{13}+\frac{1}{61}+\frac{1}{61}+\frac{1}{61}\)

A<\(\frac{1}{5}+\frac{1}{13}x3+\frac{1}{61}x3\)

A<\(\frac{1}{5}+\frac{3}{13}+\frac{3}{61}=0,4799...< \frac{1}{2}\)

Vậy A<\(\frac{1}{2}\)

Mình viết phân số lâu lắm đó tk cho mình nha. Mình cảm ơn nhiều ^-^

A bé hơn \(\frac{1}{2}\)

(1/16)¹²=(1/2^4)¹²=(1/2)⁴

(1/63)⁷>(1/64)⁷=(1/2^6)⁷=(1/2)⁴²

Có (1/2)^42>(1/2)^48

=> Kết quả là >

(1/16)¹²=(1/2^4)¹²=(1/2)⁴⁸

(1/63)⁷>(1/64)⁷=(1/2^6)⁷=(1/2)⁴²

Có (1/2)^42>(1/2)^48

=> Kết quả là >

1/63<1/16

-5<0<1/63

-101/-100=101/100>1>200/201

1/17>1/27>3/83

135/136=1+(1/136)>1+(1/137)=136/137

-371/459<0<-371/-459

267/-268>-1>-1347/1343

-13/38<-1/3<29/-88

-18/31=\(\frac{-18.10101}{31.10101}=\frac{-181818}{313131}\)

5 > 1/63

101/-100 < 200/201

1/17 > 3/83

135/136 < 136/137

-371/459 < -371/-459

267/-268 > -1347/1343

-13/38 < 29/-88

-18/31 = -181818/313131

1. ta có 298²=298.298

=> 297.299=298.298

Ta có:

\(9^{12}=\left(9^6\right)^2=531441^2\)

\(26^2=26^2\)

Vì 531441 > 26 nên 9^12>26^2

9¹² = ( 9⁶ )²

26² = 26²

9⁶ = 531441

Mà 531441 > 26 nên 9¹² > 26² 

9¹²= (9⁶)² = 531441²

Vì 531441> 26=> 531441²> 26²

Vậy 9¹² > 26²

\(\frac{13^5+13^4-13^3}{26^3}\)

\(=\frac{13^3\times(13^2+13-1)}{(2\times13)^3}\)

\(=\frac{13^3\times181}{2^3\times13^3}\)

\(=\frac{181}{8}\)

1) Tính : 

\(\frac{13^5+13^4-13^3}{26^3}\)

\(=\frac{13^3.\left(13^2+13-1\right)}{13^3.2^3}\)

\(=\frac{13^3.181}{13^3.8}\)

\(=\frac{181}{8}\)

2) Ta có : 2⁷⁵ = 2^3.25 = (2³)²⁵ = 8²⁵

                3⁵⁰ = 3^2.25 = (3²)²⁵ = 9²⁵

Vì 8²⁵ < 9²⁵ nên 2⁷⁵ < 3⁵⁰

=>  2⁷⁵ < 3⁵⁰

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

\(3^{50}=(3^2)^{25}=9^{25}\)

\(\Rightarrow8^{25}\)<\(9^{25}\)\(\Rightarrow2^{75}\)<\(3^{50}\)