($x^{2}$ -$\frac{2}{5}$y) . ($x^{2}$ -$\frac{2}{5}$y) (2x +$y^{2}

0 bình luận về “($x^{2}$ -$\frac{2}{5}$y) . ($x^{2}$ -$\frac{2}{5}$y) (2x +$y^{2}”

1. \(\begin{array}{l}
    + )\,\,\,\left( {{x^2} – \dfrac{2}{5}y} \right).\left( {{x^2} – \dfrac{2}{5}y} \right) = {\left( {{x^2} – \dfrac{2}{5}y} \right)^2}\\
    = {\left( {{x^2}} \right)^2} – 2.{x^2}.\dfrac{2}{5}y + {\left( {\dfrac{2}{5}y} \right)^2} = {x^4} – \dfrac{4}{5}{x^2}y + \dfrac{4}{{25}}{y^2}\\
    + )\,\,{\left( {2x + {y^2}} \right)^3}\\
    = {\left( {2x} \right)^3} + 3.{\left( {2x} \right)^2}.{y^2} + 3.2x.{\left( {{y^2}} \right)^2} + {\left( {{y^2}} \right)^3}\\
    = 8{x^3} + 12{x^2}{y^2} + 6x{y^4} + {y^8}\\
    + )\,\,{\left( {x + \dfrac{1}{4}} \right)^2} = {x^2} + 2.x.\dfrac{1}{4} + {\left( {\dfrac{1}{4}} \right)^2} = {x^2} + \dfrac{1}{2}x + \dfrac{1}{{16}}
   \end{array}\)

    

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