Question

$frac{ {x}^{47} + {x}^{46} + ... + x + 1 }{ {x}^{15} + {x}^{14} + ... + x + 1 }$
сократите дробь​

Answer 1

$dfrac{ {x}^{47} + {x}^{46} + ... + x + 1 }{ {x}^{15} + {x}^{14} + ... + x + 1 } =\= dfrac{ x^{32}(x^{15}+...+1)+x^{16}(x^{15}+...+1)+(x^{15}+...+1) }{ {x}^{15} + {x}^{14} + ... + x + 1 } =\=dfrac{ (x^{15}+x^{14}+...+x+1)(x^{32}+x^{16}+1) }{ {x}^{15} + {x}^{14} + ... + x + 1 } =x^{32}+x^{16}+1$