(x^2+x+1)(x^2-x+1)(x^4-x^2+1)(x^8-x^4+1) 27/07/2021 Bởi Melody (x^2+x+1)(x^2-x+1)(x^4-x^2+1)(x^8-x^4+1)
(x²+x+1)(x²-x+1)(x^4-x²+1)(x^8-x^4+1) =[(x²+1)²−x²](x^4−x²+1)(x^8−x^4+1) =(x^4+x²+1)(x^4−x²+1)(x^8−x^4+1) =[(x^4+1)2−x^4].(x^8−x^4+1) =(x^8+x^4+1)(x^8−x^4+1) =(x^8+1)²−x^8 =x^16+x^8+1 Bình luận
Đáp án: $\begin{array}{l}\left( {{x^2} + x + 1} \right)\left( {{x^2} – x + 1} \right)\left( {{x^4} – {x^2} + 1} \right)\left( {{x^8} – {x^4} + 1} \right)\\ = \left[ {{{\left( {{x^2} + 1} \right)}^2} – {x^2}} \right]\left( {{x^4} – {x^2} + 1} \right)\left( {{x^8} – {x^4} + 1} \right)\\ = \left( {{x^4} + {x^2} + 1} \right)\left( {{x^4} – {x^2} + 1} \right)\left( {{x^8} – {x^4} + 1} \right)\\ = \left[ {{{\left( {{x^4} + 1} \right)}^2} – {x^4}} \right].\left( {{x^8} – {x^4} + 1} \right)\\ = \left( {{x^8} + {x^4} + 1} \right)\left( {{x^8} – {x^4} + 1} \right)\\ = {\left( {{x^8} + 1} \right)^2} – {x^8}\\ = {x^{16}} + {x^8} + 1\end{array}$ Bình luận
(x²+x+1)(x²-x+1)(x^4-x²+1)(x^8-x^4+1)
=[(x²+1)²−x²](x^4−x²+1)(x^8−x^4+1)
=(x^4+x²+1)(x^4−x²+1)(x^8−x^4+1)
=[(x^4+1)2−x^4].(x^8−x^4+1)
=(x^8+x^4+1)(x^8−x^4+1)
=(x^8+1)²−x^8
=x^16+x^8+1
Đáp án:
$\begin{array}{l}
\left( {{x^2} + x + 1} \right)\left( {{x^2} – x + 1} \right)\left( {{x^4} – {x^2} + 1} \right)\left( {{x^8} – {x^4} + 1} \right)\\
= \left[ {{{\left( {{x^2} + 1} \right)}^2} – {x^2}} \right]\left( {{x^4} – {x^2} + 1} \right)\left( {{x^8} – {x^4} + 1} \right)\\
= \left( {{x^4} + {x^2} + 1} \right)\left( {{x^4} – {x^2} + 1} \right)\left( {{x^8} – {x^4} + 1} \right)\\
= \left[ {{{\left( {{x^4} + 1} \right)}^2} – {x^4}} \right].\left( {{x^8} – {x^4} + 1} \right)\\
= \left( {{x^8} + {x^4} + 1} \right)\left( {{x^8} – {x^4} + 1} \right)\\
= {\left( {{x^8} + 1} \right)^2} – {x^8}\\
= {x^{16}} + {x^8} + 1
\end{array}$