= 36 đó bn

tich nha!!

1*2*4+2*4*8+4*8*16+8*16*32/1*3*4+2*6*8+4*12*16+8*24*32 = 56744

1*2*4+2*4*8+4*8*16+8*16*32/1*3*4+2*6*8+4*12*16+8*24*32 = 56744

#nhớ tk 

\(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}=\dfrac{32}{1-x^{32}}\)

(2+1)(2^2+1)(2^4 +1)(2^8+1)(2^16+1) - 2^32

=1.(2+1)(2²+1)(2⁴ +1)(2⁸+1)(2¹⁶+1) - 2³²

=(2-1).(2+1)(2²+1)(2⁴ +1)(2⁸+1)(2¹⁶+1) - 2³²

=(2²-1)(2²+1)(2⁴ +1)(2⁸+1)(2¹⁶+1) - 2³²

=(2⁴-1)(2⁴ +1)(2⁸+1)(2¹⁶+1) - 2³²

=(2⁸-1)(2⁸+1)(2¹⁶+1) - 2³²

=(2¹⁶-1)(2¹⁶+1) - 2³²

=2³²-1-2³²

=-1

    (2+1 ) ( 2^2 + 1) ... (2^16 + 1) - 2^32

=  3 ( 2^2 + 1) ....( 2^16 + 1) -2^32

= ( 2^2 - 1)( 2^2 +1)....(2^16  + 1 ) - 2^32

= (2^4 - 1)( 2^4 + 1)( 2^8 + 1)( 2^16 + 1) - 2^32

= ( 2^8 - 1) ( 2^8 + 1) ( 2^16  - 1 ) - 2^32

= ( 2^ 16 - 1) (2^16 + 1) - 2^32

= 2^32 - 1 - 2^32

=-1

tìm x biết: 

câu a: (2x+1)-4(x+2)^2=9

câu b: (x+3)^2-(x-4)(x+8)=1

câu c: (x+1)^3 - (x-1)^3 - 6(x-1)^2=-19

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

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

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

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

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

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

\(\Leftrightarrow A< B\)

7) \(A=1^2-2^2+3^2-4^2+...-2004^2+2005^2\)

\(A=\left(-1\right)\left(1^{ }+2\right)+\left(-1\right)\left(3+4\right)+...+\left(-1\right)\left(2003+2004\right)+2005^2\)

\(A=-\left(1+2+3+...+2004\right)+2005^2\)

\(A=-\dfrac{2004.\left(2004+1\right)}{2}+2005^2\)

\(A=-1002.2005+2005^2\)

\(A=2005\left(2005-1002\right)=2005.1003=2011015\)

8) \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(B=\dfrac{\left(2^2-1\right)}{2-1}\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)

\(B=\left(2^{64}-1\right)-2^{64}\)

\(B=-1\)

7. 2011015

8. -1

khó quá làm sao mà trả lời đc

Vắt óc đi

tự đầu mình vắt óc mà suy nghĩ

A = (2 + 1)(2² + 1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ + 1)(2³² + 1) - 2⁶⁴

A = (2 - 1)(2 + 1)(2² + 1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ + 1)(2³² + 1) - 2⁶⁴

A = (2² - 1)(2² + 1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ + 1)(2³² + 1) - 2⁶⁴

A = (2⁴ - 1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ + 1)(2³² + 1) - 2⁶⁴

A = (2⁸ - 1)(2⁸ + 1)(2¹⁶ + 1)(2³² + 1) - 2⁶⁴

A = (2¹⁶ - 1)(2¹⁶ + 1)(2³² + 1) - 2⁶⁴

A = (2³² - 1)(2³² + 1) - 2⁶⁴

A = 2⁶⁴ - 1 - 2⁶⁴

A = -1

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

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

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

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

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

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)

=>N<M