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 \(A=2B\)

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 !

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

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

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

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

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

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

Vậy \(A< B\)

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

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

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

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

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

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

vậy B=\(3^{32}-1\)

chúc bạn hcoj tốt ^^

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

nhân A với (3-1) ta có :

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

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

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

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

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

\(=3^{32}-1\)

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

\(\Rightarrow A=B\)

A = (3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1) 

=>2A=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=(3³²-1):2

=>A<B

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xl cái kia nó bj lỗi hay sao í

B=(3+1).(3²+1).(3⁴+1).(3⁸+1).(3¹⁶+1)

 =>2B=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=\(\frac{3^{32}-1}{2}

trong sách giải có bài này