\(A=1+3+3^2+...+3^{2001}\)

\(\Rightarrow3A=3+3^2+3^3+...+3^{2002}\)

\(\Rightarrow3A-A=3+3^2+3^3+...+3^{2002}-1-3^2-3^3-...-3^{2001}\)

\(\Rightarrow2A=3^{2002}-1\)

\(\Rightarrow A=\dfrac{3^{2002}-1}{2}\)

Vì \(\dfrac{3^{2002}-1}{2}< 3^{2002}-1\Rightarrow A< B\)

a) \(\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\) 

⇔ \(\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)

⇔ \(\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}=\dfrac{x+2005}{2002}+\dfrac{x+2005}{2001}\)

⇔ \(\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)\)=0

Vì\(\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)\)<0 nên phương trinh đã cho tương đương:

x+2005=0 ⇔x=-2005

b) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\) 

⇔ \(\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)

⇔ \(\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)

⇔ \(\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)

Vì \(\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)>0\) nên phương trình đã cho tương đương:

300-x=0 ⇔ x=300

giúp mk vs

\(a=1+2+2^2+...+2^{2021}\)

\(\Rightarrow2a=2+2^2+2^3+...+2^{2022}\)

\(\Rightarrow2a-a=2+2^2+2^3+...+2^{2022}-1-2-2^2-...-2^{2021}\)

\(\Rightarrow a=2^{2022}-1\)

\(\Rightarrow a=2^{2022}-1=b\)

\(a=1+2+2^2+2^3+...+2^{2021}\)

\(2a=2+2^2+2^3+2^4...+2^{2021}+2^{2022}\)

\(2a-a=\)\(\left(2+2^2+2^3+2^4...+2^{2021}+2^{2022}\right)-\left(1+2+2^2+2^3+...+2^{2021}\right)\)

\(a=2^{2022}-1\)

⇒ a=b

a, 

A=1+3+3²+3³+3⁴+3⁵+3⁶

=> 3A=3+3²+3³+3⁴+3⁵+3⁶+3⁷

=> 3A-A=(3+3²+3³+3⁴+3⁵+3⁶+3⁷)-(1+3+3²+3³+3⁴+3⁵+3⁶)

=> 2A=3⁷-1

=> A=3⁷-1/2

Vì (3⁷-1)/2   < 3⁷-1 

=> A < B

b, C=1+2+2²+...+2²⁰⁰¹+2²⁰⁰²

=> 2C=2+2²+2³+....+2²⁰⁰²+2²⁰⁰³

=> 2C-C=(2+2²+2³+...+2²⁰⁰²+2²⁰⁰³)-(1+2+2²+...+2²⁰⁰²)

=> C=2²⁰⁰³-1

Vì 2²⁰⁰³-1 = 2²⁰⁰³-1

=> C = D.

a) \(A=1+3+3^2+...+3^6\)

\(\Rightarrow3A=3+3^2+...+3^7\)

\(\Rightarrow3A-A=3+3^2+...+3^7-1-3-3^2-...-3^6\)

\(\Rightarrow2A=3^7+2\)

\(\Rightarrow A=\frac{3^7+2}{2}\)

Vì \(3^7-1>\frac{3^7+2}{2}\)=> A < B.

b) Câu này thì nhân C cho 2 và làm tương tự như câu trên nha.

A=1+3+3^2+3^3+...+3^6

3A=3x(1+3+3^2+3^3+...+3^6)

3A-A=\(\left(3+3^2+3^3+...+3^7\right)-\left(1+3+3^2+...+3^6\right)\)

2A=3^7-1

A= \(\frac{3^7-1}{2}\)

\(\Rightarrow\)A<3^7-1 ( vì  \(\frac{3^7-1}{2}\)  <3^7-1) 

                          ( điều phải chứng minh)

C= 1+2+2^2+...+2^2001+2^2002

2C=2x( 1+2+2^2+...+2^2001+2^2002)

2C-C=(2+2^2+2^3+...+2^2002+2^2003)-( 1+2+2^2+...+2^2001+2^2002)

C=2^2003-1

\(\Rightarrow\)C=2^2003-1

              ( điều phải chứng minh)

bạn ơi bài này là bài toán dạng lũy thừa cơ bản nhất của toán nâng cao lớp 6. bạn học rồi sẽ biết.

a ) Ta có:

A = 1 + 3 + 3² + 3³+ ..... + 3⁶

A x 3 = 3 + 3² + 3³ + 3⁴ + .... + 3⁷

A x 2 - A = ( 3 + 3² + 3³ + 3⁴ + .... + 3⁷ ) - ( 1 + 3 + 3² + 3³ + .... + 3⁶ )

A = 3⁷ - 1

Mà : B = 3⁷ - 1 nên A = B

b ) Ta có :

C = 1 + 2 + 2² + 2³ + ...... + 2²⁰⁰²

C x 2 = 2 + 2² + 2³ + 2⁴ + ..... + 2²⁰⁰³

C x 2 - C = ( 2 + 2² + 2³ + 2⁴ + ...... + 2²⁰⁰³ ) - ( 1 + 2 + 2² + 2³ + ..... + 2²⁰⁰² )

C = 2²⁰⁰³ - 1 

Mà : D = 2²⁰⁰³ - 1 nên C = D

A=1+3+3^2+....+3^100
\Rightarrow 3A=3+ + +...+ 
\Rightarrow3A-A=2A=(3+ + + )-(1+3+ +....+ )
= -1
\RightarrowA=( -1):2

Bài 1: 

a) 0²⁰⁰² < 0²⁰²³

b) 2022⁰ = 2023⁰

c) 54⁹ < 55¹⁰

d) ( 4 + 5 )³ > 4² + 5²

đ) 9² - 3² > ( 9 - 3 )²

Bài 2:

a) 3² x 4³ - 3² + 333

= 9 x 64 - 9 + 333

= 576 - 9 + 333

= 567 + 333

= 900

b) 5 x 4³ + 24 x 5 + 41⁰

= 5 x 64 + 24 x 5 + 1

= 5 x ( 64 + 24 ) + 1

= 5 x 88 + 1

= 440 + 1

= 441

c) 2³ x 4² + 3² x 5 - 40 x 1²⁰²³

= 8 x 16 + 9 x 5 - 40 x 1

= 128 + 45 - 40

= 133

Bài 1 :

a) \(0^{2002}=0;0^{2023}=0\Rightarrow0^{2002}=0^{2023}\)

b) \(2022^0=1;2023^0=1\Rightarrow2022^0=2023^0\)

c) \(54^9< 55^9;55^9< 55^{10}\Rightarrow54^9< 55^{10}\)

d) \(\left(4+5\right)^3>\left(4+5\right)^2;\left(4+5\right)^2>4^2+5^2\Rightarrow\left(4+5\right)^3>4^2+5^2\)

đ) \(9^2-3^2=81-9=82;\left(9-3\right)^2=6^2=36\Rightarrow9^2-3^2>\left(9-3\right)^2\)

Bài 2 :

a) \(3^2.4^3-3^2+333=3^2\left(4^3-1\right)+9.37=9.63+9.37=9\left(63+37\right)=9.100=900\)

b) \(5.4^3+24.5+41^0=20.4^2+20.6+1=20\left(16+6\right)+1=20.22+1=441\)

c) \(2^3.4^2+3^2.5-40.1^{2023}=8.16+9.5-40.1=128+45-40=128+5=133\)

A = 1 + 2 + 2² + ... + 2^2002  

A = 1 + (2 + 2² + ... + 2^2002 )  

Ta xét :  

u1 = 2  

u2 = 2.2 = 2²  

u3 = 2.2² = 2^3  

u2002 = 2.2^2001 = 2^2002  

Tổng cấp số nhân : S = u1.(1 - q^n) / (1 - q) = 2.(1 - 2^2002) / (1 - 2) = 2(2^2002 - 1) = 2^2003 - 2  

A = 1 + 2^2003 - 2 = 2^2003 - 1  

So sánh với B  

2^2003 - 1 = 2^2003 - 1

 Vậy B = A 

A<B                      

=>2A=2+2^2+2^3+2^4+2^5+...+2^2002+2^2003

=>2A-A=2^2003-1

=>A=2^2003-1

=>A<B