Phân tích thành nhân tử: a) x^4y^4+4 b) 4x^4+1 c) 64x^4+1 d) x^4+64 e) 16x^4(x-y)-x+y f) 2x^3y-2xy^3-4xy^2-2xy g) x(y^2-z^2)+y(z^2-x^2)+z(x^2-y^2) h)

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1. Giải thích các bước giải:

   Ta có:

   \(\begin{array}{l}
   a,\\
   {x^4}{y^4} + 4 = \left( {{x^4}{y^4} + 4{x^2}{y^2} + 4} \right) – 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} + 1 = \left( {4{x^4} + 4{x^2} + 1} \right) – 4{x^2}\\
    = {\left( {2{x^2} + 1} \right)^2} – {\left( {2x} \right)^2}\\
    = \left( {2{x^2} + 1 – 2x} \right)\left( {2{x^2} + 1 + 2x} \right)\\
   c,\\
   64{x^4} + 1 = \left( {64{x^4} + 16{x^2} + 1} \right) – 16{x^2}\\
    = {\left( {8{x^2} + 1} \right)^2} – {\left( {4x} \right)^2}\\
    = \left( {8{x^2} + 1 – 4x} \right)\left( {8{x^2} + 1 + 4x} \right)\\
   d,\\
   {x^4} + 64\\
    = \left( {{x^4} + 16{x^2} + 64} \right) – 16{x^2}\\
    = {\left( {{x^2} + 8} \right)^2} – {\left( {4x} \right)^2}\\
    = \left( {{x^2} + 8 – 4x} \right)\left( {{x^2} + 8 + 4x} \right)\\
   e,\\
   16{x^4}\left( {x – y} \right) – x + y\\
    = 16{x^4}\left( {x – y} \right) – \left( {x – y} \right)\\
    = \left( {x – y} \right)\left( {16{x^4} – 1} \right)\\
    = \left( {x – y} \right)\left( {4{x^2} – 1} \right)\left( {4{x^2} + 1} \right)\\
    = \left( {x – y} \right)\left( {2x – 1} \right)\left( {2x + 1} \right)\left( {4{x^2} + 1} \right)\\
   f,\\
   2{x^3}y – 2x{y^3} – 4x{y^2} – 2xy\\
    = 2xy\left( {{x^2} – {y^2} – 2y – 1} \right)\\
    = 2xy.\left[ {{x^2} – \left( {{y^2} + 2y + 1} \right)} \right]\\
    = 2xy.\left[ {{x^2} – {{\left( {y + 1} \right)}^2}} \right]\\
    = 2xy\left( {x + y + 1} \right)\left( {x – y – 1} \right)\\
   g,\\
   x\left( {{y^2} – {z^2}} \right) + y\left( {{z^2} – {x^2}} \right) + z\left( {{x^2} – {y^2}} \right)\\
    = x\left( {y – z} \right)\left( {y + z} \right) + y{z^2} – y{x^2} + z{x^2} – z{y^2}\\
    = x\left( {y – z} \right)\left( {y + z} \right) + yz\left( {z – y} \right) – {x^2}\left( {y – z} \right)\\
    = \left( {y – z} \right)\left( {xy + xz – yz – {x^2}} \right)\\
    = \left( {y – z} \right)\left[ {y\left( {x – z} \right) + x\left( {z – x} \right)} \right]\\
    = \left( {y – z} \right)\left( {x – z} \right)\left( {y – x} \right)\\
   h,\\
   16{x^3} – 54{y^3}\\
    = 2\left( {8{x^3} – 27{y^3}} \right)\\
    = 2.\left( {2x – 3y} \right)\left( {4{x^2} + 6xy + 9{y^2}} \right)\\
   i,\\
   5{x^2} – 5{y^2} = 5\left( {{x^2} – {y^2}} \right) = 5\left( {x – y} \right)\left( {x + y} \right)\\
   k,\\
   16{x^3}y + \frac{1}{4}y{z^3}\\
    = \frac{1}{4}y\left( {64{x^3} + {z^3}} \right)\\
    = \frac{1}{4}y.\left( {4x + z} \right)\left( {16{x^2} – 4xz + {z^2}} \right)\\
   l,\\
   2{x^4} – 32 = 2.\left( {{x^4} – 16} \right) = 2.\left( {{x^2} – 4} \right)\left( {{x^2} + 4} \right) = 2\left( {x – 2} \right)\left( {x + 2} \right)\left( {{x^2} + 4} \right)
   \end{array}\)

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