B1.So sánh A=(2+1)(2^2+1)(2^4+1)(2^6+1)(2^8+1)(2^16+1) với B=2^32 B2. Cho a+b+c=0. Chứng minh rằng A^3+B^3+C^3=3abc

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1. Bài 1

   Ta có

   $A = (2 + 1)(2^2 + 1)(2^4 + 1)(2^6 + 1)(2^8 + 1)(2^{16} + 1)$

   $= 1.(2 + 1)(2^2 + 1)(2^4 + 1)(2^6 + 1)(2^8 + 1)(2^{16} + 1)$

   $= (2-1) (2 + 1)(2^2 + 1)(2^4 + 1)(2^6 + 1)(2^8 + 1)(2^{16} + 1)$

   $= (2^2-1)(2^2+1)(2^4 + 1)(2^6 + 1)(2^8 + 1)(2^{16} + 1)$

   $= (2^4-1)(2^4 + 1)(2^8+1)(2^{16} + 1)(2^6+1)$

   $= (2^8-1)(2^8+1)(2^{16} + 1)(2^6 + 1)$

   $= (2^{16} – 1)(2^{16} + 1)(2^6+1)$

   $= (2^{32}-1)(2^6 + 1)$

   Ta có

   $(2^{32} – 1)(2^6 + 1) > (2^{32} – 1). 2^6 > 2^{31} . 2^6 = 2^{37} > 2^{32}$

   Vậy $A > B$.

   Bài 2

   Ta có

   $VT = a^3 + b^3 + c^3$

   $= (a+b)^3 + c^3 – 3ab(a + b)$

   $= (a + b + c)[(a+b)^2 – c(a+b) + c^2] – 3ab(-c)$ (do $a + b + c = 0$)

   $= 3abc = VP$

   Trả lời