\(a,2n-3⋮n+1\)

\(\Rightarrow2n+2-5⋮n+1\)

\(\Rightarrow2\left(n+1\right)-5⋮n+1\)

      \(2\left(n+1\right)⋮n+1\)

\(\Rightarrow5⋮n+1\)

\(\Rightarrow n+1\inƯ\left(5\right)=\left\{-1;1;-5;5\right\}\)

\(\Rightarrow n\in\left\{-2;0;-6;4\right\}\)

vậy_

\(b,A=2^0+2^1+2^2+...+2^{100}\)

\(\Rightarrow2A=2^1+2^2+2^3+...+2^{101}\)

\(\Rightarrow2A-A=2^{101}-1\text{ hay }A=2^{101}-1\)

\(2^{101}-1< 2^{101}\)

\(\Rightarrow A< 2^{101}\)

vậy_

2n - 3 chia hết cho n+ 1 

=> 2n + 2 - 5  chia hết cho n + 1 

=> 2(n+1 ) - 5 chia hết cho n + 1 

Mà 2(n+1 ) chia hết cho n + 1  

=>  5 chia hết cho n + 1 

=> n + 1 thuộc Ư(5)= {1; -1 ; 5 ; -5 }

TH1 : n + 1 = 1 => n = 0 

TH2 : n + 1 = -1  => n = -2 

th3 : n + 1 = 5 => n = 4 

TH4 : n + 1 = -5 => n = -6 

=> n thuộc {0;-2;4;6 }

A=1+2¹+2²+2³+...+2¹⁰⁰

2A=2+2²+2³+2⁴+...+2¹⁰¹

2A-A=2¹⁰¹-1

A=2¹⁰¹-1

Ta có 2¹⁰¹>2¹⁰¹-1 nên B>A

2A=2+2^2+2^3+2^4+....+2^101

=> 2A-A=(2+2^2+2^3+2^4+....+2^101)-(1+2+2^2+2^3+...+2^100)

<=> A=2^101-1 > B=2^101

2A=2+2^2+...+2^101

=>2A-A=(2+2^2+...+2^101)-(1+2+2^2+...+2^100)

=> A=2^101-1<2^101=B

vậy a<b

Ta có \(A=1+2^2+2^3+....+2^{99}+2^{100}\)

\(2A=2+2^3+2^4+2^5+...+2^{100}+2^{101}\)

Suy ra \(2A-A=2^{101}-1=B\)

Do đó A =B

Vậy A =B 

A = 1 + 2^2 + 2^3 + ... + 2^99 + 2^100 

2A = 2 + 2^3 + 2^4 + ... + 2^100 + 2^101 

2A - A = ( 2 + 2^3 + 2^4 + ... + 2^100 + 2^101 ) - ( 1 + 2^2 + 2^3 + ... + 2^99 + 2^100 ) 

A = 2^101 - 1 

Vì A = 2^101 - 1 và B = 2^101 - 1 

=> A = B 

Vậy A=B

A=1+2^2+2^3+...+2^99+2^100

2A=2+2^3+2^4+...+2^100+2^101

2A-A=(2+2^3+2^4+...+2^100+2^101)-(1+2^2+2^3+...+2^99+2^100)

A=2^101-[2-(1+2^2)]

A=2^101-3

Vậy A=2^101-3 và B=2^101-1

=> A<B

ta có \(\frac{1}{\sqrt{x}}\)= \(\frac{2}{2\sqrt{x}}\)< \(\frac{2}{\sqrt{x}+\sqrt{x-1}}\)= 2(\(\sqrt{x}-\sqrt{x-1}\))

Áp dụng vào A \(\Rightarrow\)A < 1 + 2(\(\sqrt{2}-\sqrt{1}\)) + 2(\(\sqrt{3}-\sqrt{2}\)) + ... + 2(\(\sqrt{100}-\sqrt{99}\)) = 1 - 2 + \(2\sqrt{100}\)= \(2\sqrt{100}-1\)< \(2\sqrt{101}-1=B\)

\(\Rightarrow\)A < B

2:

a: A=1+2+2^2+2^3+2^4

=>2A=2+2^2+2^3+2^4+2^5

=>A=2^5-1

=>A=B

b: C=3+3^2+...+3^100

=>3C=3^2+3^3+...+3^101

=>2C=3^101-3

=>\(C=\dfrac{3^{101}-3}{2}\)

=>C=D

Ta có: 

\(\left\{\begin{matrix}5^{27}=\left(5^3\right)^9=125^9\\2^{63}=\left(2^7\right)^9=128^9\end{matrix}\right\}\Rightarrow5^{27}< 2^{63}\left(1\right)\)

\(\left\{\begin{matrix}2^{63}=\left(2^9\right)^7=512^7\\5^{28}=\left(5^4\right)^7=625^7\end{matrix}\right\}\Rightarrow2^{63}< 5^{28}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow5^{27}< 2^{63}< 5^{28}\) (đpcm)

 \(a.5^{27}=\left(5^3\right)^9=125^9\\ 2^{63}=\left(2^7\right)^9=128^9\)

Vì 128⁹ > 125⁹ => 2⁶³ > 5²⁷

^{\(5^{28}=\left(5^4\right)^7=625^7\\ 2^{63}=\left(2^9\right)^7=512^7\)}

Vì 625⁷ > 512⁷ = > 5²⁸ > 2⁶³

Đã CMR: \(5^{27}< 2^{63}< 5^{28}\)

\(b.A=1+2+2^2+2^3+2^4\\ 2A=2+2^2+2^3+2^4+2^5\\ 2A-A=\left(2+2^2+2^3+2^4+2^5\right)-\left(1+2+2^2+2^3+2^4+\right)\\ A=2^5-1\\ 2^5-1=2^5-1=>A=B\\ c,C=3+3^2+....+3^{100}\\ 3C=3^2+......+3^{101}\\ 3C-C=\left(3^2+...+3^{101}\right)-\left(3+...+3^{100}\right)\\ 2C=3^{101}-3\\ C=\dfrac{3^{101}-3}{2}\\ \dfrac{3^{101}-3}{2}=\dfrac{3^{101}-3}{2}=>C=D\)

Hong bé ơi.Bé hong follow anh mà đòi xin đáp án của anh à

ko

mình nghĩ là B