Phân tích đa thức thành nhân tử: a,x^4+2x^3-6x-9 b,(a^2+b^2+ab)^2-a^2b^2-b^2c^2-c^2a^2 c,x^2(y^2-4)-6y(y^2-4)+9 d,a^4-9a^3+81a-81

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1. Giải thích các bước giải:

   $\begin{array}{l}
   a){x^4} + 2{x^3} – 6x – 9\\
    = \left( {{x^4} – 9} \right) + \left( {2{x^3} – 6x} \right)\\
    = \left( {{x^2} – 3} \right)\left( {{x^2} + 3} \right) + 2x\left( {{x^2} – 3} \right)\\
    = \left( {{x^2} – 3} \right)\left( {{x^2} + 3 + 2x} \right)\\
    = \left( {{x^2} – 3} \right)\left( {{x^2} + 2x + 3} \right)\\
   b){\left( {{a^2} + {b^2} + ab} \right)^2} – {a^2}{b^2} – {b^2}{c^2} – {c^2}{a^2}\\
    = {\left( {{a^2} + {b^2} + ab} \right)^2} – {\left( {ab} \right)^2} – {c^2}\left( {{b^2} + {a^2}} \right)\\
    = \left( {{a^2} + {b^2} + ab – ab} \right)\left( {{a^2} + {b^2} + ab + ab} \right) – {c^2}\left( {{b^2} + {a^2}} \right)\\
    = \left( {{a^2} + {b^2}} \right)\left( {{a^2} + 2ab + {b^2}} \right) – {c^2}\left( {{a^2} + {b^2}} \right)\\
    = \left( {{a^2} + {b^2}} \right){\left( {a + b} \right)^2} – {c^2}\left( {{b^2} + {a^2}} \right)\\
    = \left( {{a^2} + {b^2}} \right)\left( {{{\left( {a + b} \right)}^2} – {c^2}} \right)\\
    = \left( {{a^2} + {b^2}} \right)\left( {a + b – c} \right)\left( {a + b + c} \right)\\
   c){x^2}\left( {{y^2} – 4} \right) – 6y\left( {{y^2} – 4} \right) + 9\\
    = \left( {{y^2} – 4} \right)\left( {{x^2} – 6y} \right) + 9\\
   d){a^4} – 9{a^3} + 81a – 81\\
    = \left( {{a^4} – 81} \right) – 9a\left( {{a^2} – 9} \right)\\
    = \left( {{a^2} – 9} \right)\left( {{a^2} + 9} \right) – 9a\left( {{a^2} – 9} \right)\\
    = \left( {{a^2} – 9} \right)\left( {{a^2} + 9 – 9a} \right)\\
    = \left( {{a^2} – 9} \right)\left( {{a^2} – 9a + 9} \right)
   \end{array}$

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2. Đáp án:

   a/ $(x-\sqrt{3})(x+\sqrt{3})(x^2+2x+3)$

   b/ $(a^2+b^2)(a+b-c)(a+b+c)$

   c/ $(a^2+b^2)(a+b-c)(a+b+c)$

   d/ $(a-3)(a+3)(a^2-9a+9)$

   Giải thích các bước giải:

   a/ $x^4+2x^3-6x-9$

   $=(x^4-9)+(2x^3-6x)$

   $=(x^2-3)(x^2+3)+2x(x^2-3)$

   $=(x^2-3)(x^2+3+2x)$

   $=(x-\sqrt{3})(x+\sqrt{3})(x^2+2x+3)$

   b/ $(a^2+b^2+ab)^2-a^2b^2-b^2c^2-c^2a^2$

   $=a^4+b^4+a^2b^2+2a^2b^2+2ab^3+2a^3b-a^2b^2-b^2c^2-c^2a^2$

   $=a^4+b^4+2a^2b^2+2ab^3+2a^3b-b^2c^2-c^2a^2$

   $=(a^2+b^2)^2+2ab(a^2+b^2)-c^2(a^2+b^2)$

   $=(a^2+b^2)(a^2+b^2+2ab-c^2)$

   $=(a^2+b^2).[(a+b)^2-c^2]$

   $=(a^2+b^2)(a+b-c)(a+b+c)$

   d/ $a^4-9a^3+81a-81$

   $=(a^4-81)-(9a^3-81a)$

   $=(a^2-9)(a^2+9)-9a(a^2-9)$

   $=(a^2-9)(a^2+9-9a)$

   $=(a-3)(a+3)(a^2-9a+9)$

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