g) x^5 +x^4 +1 h) x^5 +x +1 i) x^8 +x^7 +1 k) x^5 – x^4 -1 l) x^7 + x^5 +1 m) x^8 +x^4 +1 lam ra giay part2 nha 30/09/2021 Bởi Abigail g) x^5 +x^4 +1 h) x^5 +x +1 i) x^8 +x^7 +1 k) x^5 – x^4 -1 l) x^7 + x^5 +1 m) x^8 +x^4 +1 lam ra giay part2 nha
\[\begin{array}{l} g)\,\,\,{x^5} + {x^4} + 1 = {x^5} – {x^2} + {x^4} + {x^2} + 1\\ = {x^2}\left( {{x^3} – 1} \right) + \left( {{x^4} + {x^2} + 1} \right)\\ = {x^2}\left( {x – 1} \right)\left( {{x^2} + x + 1} \right) + \left( {{x^4} + 2{x^2} + 1 – {x^2}} \right)\\ = {x^2}\left( {x – 1} \right)\left( {{x^2} + x + 1} \right) + {\left( {{x^2} + 1} \right)^2} – {x^2}\\ = {x^2}\left( {x – 1} \right)\left( {{x^2} + x + 1} \right) + \left( {{x^2} – x + 1} \right)\left( {{x^2} + x + 1} \right)\\ = \left( {{x^2} + x + 1} \right)\left( {{x^3} – {x^2} + {x^2} – x + 1} \right)\\ = \left( {{x^2} + x + 1} \right)\left( {{x^3} – x + 1} \right).\\ h)\,\,{x^5} + x + 1 = {x^5} – {x^2} + {x^2} + x + 1\\ = {x^2}\left( {{x^3} – 1} \right) + {x^2} + x + 1\\ = {x^2}\left( {x – 1} \right)\left( {{x^2} + x + 1} \right) + {x^2} + x + 1\\ = \left( {{x^2} + x + 1} \right)\left( {{x^3} – {x^2} + 1} \right).\\ m)\,\,{x^8} + {x^4} + 1 = {\left( {{x^4}} \right)^2} + 2{x^4} + 1 – 2{x^4}\\ = {\left( {{x^4} + 1} \right)^2} – 2{x^4} = \left( {{x^4} + 1 – \sqrt 2 {x^2}} \right)\left( {{x^4} + 1 + \sqrt 2 {x^2}} \right). \end{array}\] Bình luận
Đáp án: g)x^5+x^4+1=x^5-x^2+x^4-x+x^2+x+1=x^2(x-1)(x^2+x+1)+x(x-1)(x^2+x+1)+x^2+x+1=(x^2+x+1)(x^3-x^2+x^2-x+1)=(x^2+x+1)(x^3-x+1) h) x^5+x+1=x^5-x^2+x^2+x+1=x^2(x-1)(x^2+x+1)+(x^2+x+1)=(x^2+x+1)(x^3-x^2+1) i)x^8+x^7+1=x^8-x^2+x^7-x+x^2+x+1=x^2(x^6-1)+x(x^6-1)+x^2+x+1=x^2(x^3+1)(x-1)(x^2+x+1)+x(x^3+1)(x-1)(x^2+x+1)+(x^2+x+1)=(x^2+x+1)(x^6+x^3-x^4-x+1) k) x^5-x^4-1=x^5+x^2-x^4-x-x^2+x-1=x^2(x+1)(x^2-x+1)-x(x+1)(x^2-x+1)-(x^2-x+1)=(x^2-x+1)(x^3+x^2-x^2-x-1)=(x^2-x+1)(x^3-x-1) l)x^7+x^5+1=x^7-x+x^5-x^2+x^2+x+1=x(x^3+1)(x^3-1)+x^2(x^3-1)+(x^2+x+1)=x(x^3+1)(x-1)(x^2+x+1)+x^2(x-1)(x^2+x+1)+(x^2+x+1)=(x^2+x+1)(x^5-x^4+x^3-x+1) m)x^8+x^4+1=x^8-x^2+x^4-x+x^2+x+1=x^2(x^3+1)(x^3-1)+x(x^3-1)+x^2+x+1=x^2(x^3+1)(x-1)(x^2+x+1)+x(x-1)(x^2+x+1)+(x^2+x+1)=(x^2+x+1)(x^6-x^5+x^3-x+1) Bình luận
\[\begin{array}{l}
g)\,\,\,{x^5} + {x^4} + 1 = {x^5} – {x^2} + {x^4} + {x^2} + 1\\
= {x^2}\left( {{x^3} – 1} \right) + \left( {{x^4} + {x^2} + 1} \right)\\
= {x^2}\left( {x – 1} \right)\left( {{x^2} + x + 1} \right) + \left( {{x^4} + 2{x^2} + 1 – {x^2}} \right)\\
= {x^2}\left( {x – 1} \right)\left( {{x^2} + x + 1} \right) + {\left( {{x^2} + 1} \right)^2} – {x^2}\\
= {x^2}\left( {x – 1} \right)\left( {{x^2} + x + 1} \right) + \left( {{x^2} – x + 1} \right)\left( {{x^2} + x + 1} \right)\\
= \left( {{x^2} + x + 1} \right)\left( {{x^3} – {x^2} + {x^2} – x + 1} \right)\\
= \left( {{x^2} + x + 1} \right)\left( {{x^3} – x + 1} \right).\\
h)\,\,{x^5} + x + 1 = {x^5} – {x^2} + {x^2} + x + 1\\
= {x^2}\left( {{x^3} – 1} \right) + {x^2} + x + 1\\
= {x^2}\left( {x – 1} \right)\left( {{x^2} + x + 1} \right) + {x^2} + x + 1\\
= \left( {{x^2} + x + 1} \right)\left( {{x^3} – {x^2} + 1} \right).\\
m)\,\,{x^8} + {x^4} + 1 = {\left( {{x^4}} \right)^2} + 2{x^4} + 1 – 2{x^4}\\
= {\left( {{x^4} + 1} \right)^2} – 2{x^4} = \left( {{x^4} + 1 – \sqrt 2 {x^2}} \right)\left( {{x^4} + 1 + \sqrt 2 {x^2}} \right).
\end{array}\]
Đáp án: g)x^5+x^4+1=x^5-x^2+x^4-x+x^2+x+1=x^2(x-1)(x^2+x+1)+x(x-1)(x^2+x+1)+x^2+x+1=(x^2+x+1)(x^3-x^2+x^2-x+1)=(x^2+x+1)(x^3-x+1)
h) x^5+x+1=x^5-x^2+x^2+x+1=x^2(x-1)(x^2+x+1)+(x^2+x+1)=(x^2+x+1)(x^3-x^2+1)
i)x^8+x^7+1=x^8-x^2+x^7-x+x^2+x+1=x^2(x^6-1)+x(x^6-1)+x^2+x+1=x^2(x^3+1)(x-1)(x^2+x+1)+x(x^3+1)(x-1)(x^2+x+1)+(x^2+x+1)=(x^2+x+1)(x^6+x^3-x^4-x+1)
k) x^5-x^4-1=x^5+x^2-x^4-x-x^2+x-1=x^2(x+1)(x^2-x+1)-x(x+1)(x^2-x+1)-(x^2-x+1)=(x^2-x+1)(x^3+x^2-x^2-x-1)=(x^2-x+1)(x^3-x-1)
l)x^7+x^5+1=x^7-x+x^5-x^2+x^2+x+1=x(x^3+1)(x^3-1)+x^2(x^3-1)+(x^2+x+1)=x(x^3+1)(x-1)(x^2+x+1)+x^2(x-1)(x^2+x+1)+(x^2+x+1)=(x^2+x+1)(x^5-x^4+x^3-x+1)
m)x^8+x^4+1=x^8-x^2+x^4-x+x^2+x+1=x^2(x^3+1)(x^3-1)+x(x^3-1)+x^2+x+1=x^2(x^3+1)(x-1)(x^2+x+1)+x(x-1)(x^2+x+1)+(x^2+x+1)=(x^2+x+1)(x^6-x^5+x^3-x+1)