\(∘backwin\)

\(a ) ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750\)

\( ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 5750 \)

\( 100 x + ( 1 + 100 ) ×100 : 2 = 5750\)

\(100 x + 5050 = 5750\)

\( 100 x = 5750 − 5050\)

\(100 x = 700\)

\(x = 700 : 100\)

\(x = 7\)

\(b,\) \(B=\)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2020}+2021\)

\( B < 1 -\)\(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\)

\(B<1-\)\(\dfrac{1}{2021}\)

\(B<\)\(\dfrac{2020}{2021}\)

\(\dfrac{2020}{2021}< 1\)

\(B<1\)

a) (x+1) +(x+2 ) + ...+(x+100)=5750
= 100x + (1+2+3+...+100) = 5750
=100x + 5050 = 5750
--> 100x = 5750-5050=700
--> x=7

b) Ta thấy: 1/2^2 < 1/2.3
                  1/3^2 < 1/3.4
                        ...
                  1/2021^2 < 1/2021.2022
--> B=1/2^2 + 1/3^2 + 1/4^2 + ...+ 1/2021^2 < 1/2.3 + 1/3.4 + ... +1/2021.2022 (1)
     Ta có: 1/2.3 + 1/3.4 + ... +1/2021.2022
         =1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2021 - 1/2022
         =1/2 - 1/2022 < 1 (2)
Từ (1) và (2) --> B<1 (đpcm)
                                                                      < 
                                                                     

B=(3+1).(3^2+1).(3^4+1).(3^8+1).(3^16 +1).2

=>B=2.(3+1)(3²+1)(3⁴+1)(3⁸+1).(3¹⁶+1)

=(3-1)(3+1)(3²+1)(3⁴+1)(3⁸+1).(3¹⁶+1)

=(3²-1)(3²+1)(3⁴+1)(3⁸+1).(3¹⁶+1)

=(3⁴-1)(3⁴+1)(3⁸+1).(3¹⁶+1)

=(3⁸-1)(3⁸+1).(3¹⁶+1)

=(3¹⁶-1)(3¹⁶+1)

=3³²-1=A

Vậy A=B

\(B=2.\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right).\left(3^{16}+1\right)\)

    \(=\left(3-1\right).\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

     \(=\left(3^2-1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

     \(=\left(3^4-1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

     \(=\left(3^8-1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

     \(=\left(3^{16}-1\right).\left(3^{16}+1\right)\)

     \(=3^{32}-1\)

Vậy A = B = 3³² - 1

ta có A=3³²-1=(3¹⁶+1)(3¹⁶-1)=(3¹⁶+1)(3⁸+1)(3⁸-1)=(3¹⁶+1)(3⁸+1)(3⁴+1)(3⁴-1)

=(3¹⁶+1)(3⁸+1)(3⁴+1)(3²+1)(3²-1)=(3¹⁶+1)(3⁸+1)(3⁴+1)(3²+1)(3+1)(3-1)=B

=> A=B

\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(2B=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(2B=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(2B=\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(B=\dfrac{3^{32}-1}{2}< A=3^{32}-1\)

\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =>2B=2.\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^{16}-1\right)\left(3^{16}+1\right)=3^{32}-1\\ =>A=\dfrac{3^{32}-1}{2}< B\)

\(B=\left(3+1\right)\left(3^2+1\right)...\left(3^{16}+1\right)=\dfrac{\left(3-1\right)\left(3+1\right)...\left(3^{16}+1\right)}{2}=\dfrac{\left(3^2-1\right)\left(3^2+1\right)...\left(3^{16}+1\right)}{2}=\dfrac{\left(3^4-1\right)\left(3^4+1\right)...\left(3^{16}+1\right)}{2}=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^8+16\right)}{2}=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2}=\dfrac{3^{32}-1}{2}=\dfrac{A}{2}\)

Vậy \(A=2B\)

\(A=4.\left(3^2+1\right).\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(=\frac{3^{32}-1}{2}< 3^{32}-1=B\)

Vậy \(A< B\)

Nó hơi dài cậu chờ tí nka !

Mình ghi nhầm đề bài 1 tí đề bài là :

So sánh 2 số A và B biết : 

A = (3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1) và B = 3^32 - 1

A = (2-1)(2+1)(2^2 + 1 ) (2^4 + 1 ) ( 2^8 + 1) ( 2^16 + 1)

A = (2^2 - 1)(2^2 + 1 ) ( 2^4 + 1 )(2^8 + 1 )(2^16 + 1) 

A= ( 2^4 - 1 )( 2^4 + 1 )(2^8 + 1 )(2^16 + 1 )

A = (2^8 - 1 )(2^8 + 1 )(2^16 + 1 )

A = (2^16 - 1 )(2^16 + 1 )

A =  2^32 - 1 < 2^32 = B
Vậy A = B
k mik nka !

A*2=(3-1)*(3+1)*(3^2+1)*....*(3^16+1)

A*2=(3^2-1)*(3^2+1)*(3^4+1)....*(3^16+1)

A*2=((3^4)^2-1^2)*(3^4+1)......*(3*16+1)

2*A=(3^8-1)*...(3^16+1)

bạn lm tiếp nha

nhân vào A 3^2-1

\(B=3^{32}-1=\left(3^{16}+1\right)\left(3^{16}-1\right)\)

\(=\left(3^{16}+1\right)\left(3^8+1\right)\left(3^8-1\right)\)

\(=\)\(\left(3^{16}+1\right)\left(3^8+1\right)\left(3^4+1\right)\left(3^4-1\right)\)

\(=\left(3^{16}+1\right)\left(3^8+1\right)\left(3^{\text{4}}+1\right)\left(3^2+1\right)\left(3^2-1\right)\)

\(=\left(3^{16}+1\right)\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\left(3-1\right)\)

\(=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(B=2A\)

đúng 100% k nha