1, A = 2⁹¹ = 2^7.13 = (2¹³)⁷ = 8192⁷

B = 5³⁵ = 5^5.7 = (5⁵)⁷ = 3125⁷

Vì 3125 < 8192

=> 3125⁷ < 8192⁷

=> B < A

2.Ta có:

 A=11+11²+11³+11⁴+...+11¹⁹⁹+11²⁰⁰.

11A=11²+11³+11⁴+...+11¹⁹⁹+11²⁰⁰+11²⁰¹.

11A-A=11²⁰¹-11.

10A=11²⁰¹-11.

A=(11²⁰¹-11):10

Quan sát 2 vế A và B thì ta thấy rõ ràng vế A<B hay B>A.

a)

Vì \(\frac{2009}{2010}< 1\Rightarrow\frac{2009}{2010}< \frac{2009+1}{2010+1}=\frac{2010}{2011}\)

Cần nhớ:

Nếu: \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+n}{b+n}\left(n\inℕ^∗\right)\)

Và tương tự:  \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{b+n}\left(n\inℕ^∗\right)\)

b)Ta có:

 \(\frac{1}{3^{400}}=\frac{1}{\left(3^4\right)^{100}}=\frac{1}{81^{100}}\)

\(\frac{1}{4^{300}}=\frac{1}{\left(4^3\right)^{100}}=\frac{1}{64^{100}}\)

Vì: \(81^{100}>64^{100}\Leftrightarrow\frac{1}{81^{100}}< \frac{1}{64^{100}}\Leftrightarrow\frac{1}{3^{400}}< \frac{1}{4^{300}}\)

c) Ta có:

\(\frac{200+201}{201+202}=\frac{401}{403}< 1\)

\(\frac{200}{201}+\frac{201}{202}=1-\frac{1}{201}+1-\frac{1}{202}=2-\left(\frac{1}{201}+\frac{1}{202}\right)>1\)

=>\(\frac{200}{201}+\frac{201}{202}>\frac{200+201}{201+202}\)

a, \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{200}-1\right)\)

\(-A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{200}\right)\)

\(-A=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{199}{200}\)

\(-A=\frac{1}{200}\)

\(A=\frac{-1}{200}>\frac{-1}{199}\)

Bài 1:

a: Sửa đề: 1/3^200

1/2^300=(1/8)^100

1/3^200=(1/9)^100

mà 1/8>1/9

nên 1/2^300>1/3^200

b: 1/5^199>1/5^200=1/25^100

1/3^300=1/27^100

mà 25^100<27^100

nên 1/5^199>1/3^300

Ta có p = [ ( 3 + 3² + 3³ + ... + 3²⁰¹ ) - ( 1 + 3 + 3² + ... + 3²⁰⁰ ) ] : 2

=> p = ( 3²⁰¹ - 1 ) : 2

Vì ( 3²⁰¹ - 1 ) : 2 < 3²⁰¹ nên p < Q

Ta có :

\(P=1+3+3^2+3^3+...+3^{200}\)

=> \(3P=3+3^2+3^3+3^4+...+3^{201}\)

=> \(3P-P=\left(3+3^2+3^3+...+3^{201}\right)-\left(1+3+3^2+...+3^{200}\right)\)

=> \(2P=3^{201}-1\)

Ta có :  3²⁰¹ - 1 < 3²⁰¹  =>  2P < Q => P < Q