1/ (8a³-27b³ )-2a (4a²-2b²) 2/ (3x-1)² -16 3/ (3x+1)²-4(x-2)² 4/1-(x²-2xy+y²)²

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1. Đáp án:

   $\begin{array}{l}
   1)\left( {8{a^3} – 27{b^3}} \right) – 2a\left( {4{a^2} – 2{b^2}} \right)\\
    = 8{a^3} – 27{b^3} – 8{a^3} + 4a{b^2}\\
    = 4a{b^2} – 27{b^3}\\
    = {b^2}\left( {4a – 27b} \right)\\
   2){\left( {3x – 1} \right)^2} – 16\\
    = {\left( {3x – 1} \right)^2} – {4^2}\\
    = \left( {3x – 1 – 4} \right)\left( {3x – 1 + 4} \right)\\
    = \left( {3x – 5} \right)\left( {3x + 3} \right)\\
   3){\left( {3x + 1} \right)^2} – 4{\left( {x – 2} \right)^2}\\
    = \left( {3x + 1 – 2\left( {x – 2} \right)} \right)\left( {3x + 1 + 2\left( {x – 2} \right)} \right)\\
    = \left( {x + 5} \right)\left( {5x – 3} \right)\\
   4)1 – {\left( {{x^2} – 2xy + {y^2}} \right)^2}\\
    = \left( {1 – {x^2} + 2xy – {y^2}} \right)\left( {1 + {x^2} – 2xy + {y^2}} \right)\\
    = \left( {{x^2} – 2xy + {y^2} + 1} \right)\left( {1 – {{\left( {x – y} \right)}^2}} \right)\\
    = \left( {{x^2} – 2xy + {y^2} + 1} \right)\left( {1 – x + y} \right)\left( {1 + x – y} \right)
   \end{array}$

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