a. x^2-2x-y^2+1 b. 4x^2+4x-9y^2+4 c. 5x^3-20x^2+5x d. x^2(x-3)+12-4x e. 4x^4+y^4 g. x^4+4+4x^2-4x^2 h. 6x^2-12x-18 i. x^4+5x^3-8x-40 k. x^2

0 bình luận về “a. x^2-2x-y^2+1 b. 4x^2+4x-9y^2+4 c. 5x^3-20x^2+5x d. x^2(x-3)+12-4x e. 4x^4+y^4 g. x^4+4+4x^2-4x^2 h. 6x^2-12x-18 i. x^4+5x^3-8x-40 k. x^2”

1. Giải thích các bước giải:

    Ta có:

   \(\begin{array}{l}
   a,\\
   {x^2} – 2x – {y^2} + 1 = \left( {{x^2} – 2x + 1} \right) – {y^2} = {\left( {x – 1} \right)^2} – {y^2}\\
    = \left( {x – 1 + y} \right)\left( {x – 1 – y} \right)\\
   b,\\
   4{x^2} + 4x – 9{y^2} + 1 = \left( {4{x^2} + 4x + 1} \right) – 9{y^2}\\
    = {\left( {2x + 1} \right)^2} – {\left( {3y} \right)^2} = \left( {2x + 1 – 3y} \right)\left( {2x + 1 + 3y} \right)\\
   c,\\
   5{x^3} – 20{x^2} + 5x = 5x\left( {{x^2} – 4x + 1} \right)\\
   d,\\
   {x^2}\left( {x – 3} \right) + 12 – 4x = {x^2}\left( {x – 3} \right) – 4.\left( {x – 3} \right)\\
    = \left( {x – 3} \right)\left( {{x^2} – 4} \right) = \left( {x – 3} \right)\left( {x – 2} \right)\left( {x + 2} \right)\\
   e,\\
   4{x^4} + {y^4} = \left( {4{x^4} + 4{x^2}{y^2} + {y^4}} \right) – 4{x^2}{y^2}\\
    = {\left( {2{x^2} + {y^2}} \right)^2} – {\left( {2xy} \right)^2} = \left( {2{x^2} – 2xy + {y^2}} \right)\left( {2{x^2} + 2xy + {y^2}} \right)\\
   g,\\
   {x^4} + 4 + 4{x^2} – 4{x^2} = \left( {{x^4} + 4{x^2} + 4} \right) – 4{x^2}\\
    = {\left( {{x^2} + 2} \right)^2} – {\left( {2x} \right)^2} = \left( {{x^2} – 2x + 2} \right)\left( {{x^2} + 2x + 2} \right)\\
   h,\\
   6{x^2} – 12x – 18 = 6.\left( {{x^2} – 2x – 3} \right) = 6.\left[ {\left( {{x^2} – 3x} \right) + \left( {x – 3} \right)} \right]\\
    = 6.\left[ {x\left( {x – 3} \right) + \left( {x – 3} \right)} \right] = 6\left( {x – 3} \right)\left( {x + 1} \right)\\
   i,\\
   {x^4} + 5{x^3} – 8x – 40 = \left( {{x^4} + 5{x^3}} \right) – \left( {8x + 40} \right)\\
    = {x^3}\left( {x + 5} \right) – 8\left( {x + 5} \right) = \left( {x + 5} \right)\left( {{x^3} – 8} \right)\\
    = \left( {x + 5} \right)\left( {x – 2} \right)\left( {{x^2} + 2x + 4} \right)\\
   k,\\
   {x^2} – {y^2} + {z^2} – {t^2} – 2xz + 2yt\\
    = \left( {{x^2} – 2xz + {z^2}} \right) – \left( {{y^2} – 2yt + {t^2}} \right)\\
    = {\left( {x – z} \right)^2} – {\left( {y – t} \right)^2}\\
    = \left( {x – z + y – t} \right)\left( {x – z – y + t} \right)\\
   m,\\
   {x^2} – 9x + 8 = \left( {{x^2} – 8x} \right) – \left( {x – 8} \right) = x\left( {x – 8} \right) – \left( {x – 8} \right) = \left( {x – 8} \right)\left( {x – 1} \right)
   \end{array}\)

   Bình luận