Tính giá trị biểu thức: `\text{A =}` $(1-x)(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})+x^{32}$

2 bình luận về “Tính giá trị biểu thức: `\text{A =}` $(1-x)(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})+x^{32}$”

1. Giải đáp:

   A=(1-x).(1+x).(1+x^{2}).(1+x^{4}).(1+x^{8}).(1+x^{16})+x^{32}

   Ta áp dụng HĐT: (A-B).(A+B)=A^{2}-B^{2}

   =(1^{2}-x^{2}).(1+x^{2}).(1+x^{4}).(1+x^{8}).(1+x^{16})+x^{32}

   =(1-x^{2}).(1+x^{2}).(1+x^{4}).(1+x^{8}).(1+x^{16})+x^{32}

   =[1^{2}-(x^{2})^{2}].(1+x^{4}).(1+x^{8}).(1+x^{16})+x^{32}

   =(1-x^{4}).(1+x^{4}).(1+x^{8}).(1+x^{16})+x^{32}

   =[1-(x^{4})^{2}].(1+x^{8}).(1+x^{16})+x^{32}

   =(1-x^{8}).(1+x^{8}).(1+x^{16})+x^{32}

   =[1^{2}-(x^{8})^{2}].(1+x^{16})+x^{32}

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

   =[1^{2}-(x^{16})^{2}]+x^{32}

   =1-x^{32}+x^{32}

   =1

   Vậy A=1

    

   Trả lời
2. A=(1-x).(1+x).(1+x^2).(1+x^4).(1+x^8).(1+x^{16})+x^{32}

   A=(1-x^2).(1+x^2).(1+x^4).(1+x^8).(1+x^{16})+x^{32}

   A=(1-x^4).(1+x^4).(1+x^8).(1+x^{16})+x^{32}

   A=(1-x^8).(1+x^8).(1+x^{16})+x^{32}

   A=(1-x^{16}).(1+x^{16})+x^{32}

   A=1-x^{32}+x^{32}

   A=1

   Vậy A=1

   $\\$

   **(1-x^4).(1+x^4)=1^2-(x^4)^2=1-x^8

       (1-x^8).(1+x^8)=1^2-(x^8)^2=1-x^{16}

       (1-x^{16}).(1+x^{16})=1^2-(x^{16})^2=1-x^{32}

   Trả lời