Toán Lớp 8: Phân tích đa thức thành nhân tử
a) x^4y^4+4
b) 4x^4y^4+1
c) x^4+1
d) x^8+x+1
e) x^8+x^7+1
f) x^8+3x^4+1
g) x^4+4y^4
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Toán Lớp 8: Phân tích đa thức thành nhân tử
a) x^4y^4+4
b) 4x^4y^4+1
c) x^4+1
d) x^8+x+1
e) x^8+x^7+1
f) x^8+3x^4+1
g) x^4+4y^4
Comments ( 1 )
a){x^4}{y^4} + 4\\
= {x^4}{y^4} + 4{x^2}{y^2} + 4 – 4{x^2}{y^2}\\
= {\left( {{x^2}{y^2} + 2} \right)^2} – {\left( {2xy} \right)^2}\\
= \left( {{x^2}{y^2} + 2 + 2xy} \right)\left( {{x^2}{y^2} + 2 – 2xy} \right)\\
b)4{x^4}{y^4} + 1\\
= 4{x^4}{y^4} + 4{x^2}{y^2} + 1 – 4{x^2}{y^2}\\
= {\left( {2{x^2}{y^2} + 1} \right)^2} – {\left( {2xy} \right)^2}\\
= \left( {2{x^2}{y^2} + 2xy + 1} \right)\left( {2{x^2}{y^2} – 2xy + 1} \right)\\
c){x^4} + 1\\
= {x^4} + 2{x^2} + 1 – 2{x^2}\\
= {\left( {{x^2} + 1} \right)^2} – {\left( {\sqrt 2 x} \right)^2}\\
= \left( {{x^2} + 1 + \sqrt 2 x} \right)\left( {{x^2} + 1 – \sqrt 2 x} \right)\\
d){x^8} + x + 1\\
= {x^8} + {x^7} – {x^7} + {x^6} – {x^6} + {x^5} – {x^5} + {x^4} – {x^4}\\
+ {x^3} – {x^3} + {x^2} – {x^2} + x + 1\\
= \left( {{x^8} + {x^7} + {x^6}} \right) – \left( {{x^7} + {x^6} + {x^5}} \right)\\
+ \left( {{x^5} + {x^4} + {x^3}} \right) – \left( {{x^4} + {x^3} + {x^2}} \right)\\
+ {x^2} + x + 1\\
= \left( {{x^2} + x + 1} \right)\left( {{x^6} – {x^5} + {x^3} – {x^2} + 1} \right)\\
e){x^8} + {x^7} + 1\\
= {x^8} + {x^7} – {x^6} + {x^6} + 1\\
= {x^6}\left( {{x^2} + x + 1} \right) – \left( {{x^6} – 1} \right)\\
= {x^6}\left( {{x^2} + x + 1} \right) – \left( {{x^3} – 1} \right)\left( {{x^3} + 1} \right)\\
= {x^6}\left( {{x^2} + x + 1} \right) – \left( {{x^3} + 1} \right)\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)\\
= \left( {{x^2} + x + 1} \right)\left( {{x^6} – \left( {x – 1} \right)\left( {{x^3} + 1} \right)} \right)\\
= \left( {{x^2} + x + 1} \right)\left( {{x^6} – {x^4} + {x^3} – x – 1} \right)\\
f){x^8} + 3{x^4} + 1\\
= {x^8} + 2.{x^4}.\dfrac{3}{2} + \dfrac{9}{4} – \dfrac{5}{4}\\
= {\left( {{x^4} + \dfrac{3}{2}} \right)^2} – \dfrac{5}{4}\\
= \left( {{x^4} + \dfrac{{3 – \sqrt 5 }}{2}} \right)\left( {{x^4} + \dfrac{{3 + \sqrt 5 }}{2}} \right)\\
g){x^4} + 4{y^4}\\
= {x^4} + 4{x^2}{y^2} + 4{y^4} – 4{x^2}{y^2}\\
= {\left( {{x^2} + 2{y^2}} \right)^2} – {\left( {2xy} \right)^2}\\
= \left( {{x^2} + 2{y^2} + 2xy} \right)\left( {{x^2} + 2{y^2} – 2xy} \right)
\end{array}$