Tính: a) (x/2-y)^3 b) (x/2+y/3)^3 c) (2x/3-2y)^3 d) (x+y)^3+(x-y)^3

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1. `a) (x/2-y)^3 = (x/2)^3 – 3(x/2)^2y + 3x/2y^2 – y^3`

                        `= {x^3}/8 – 3{x^2}/4y + 3x/2y^2 – y^3`

   `b) (x/2+y/3)^3 = (x/2)^3 + 3(x/2)^2y/3 + 3x/2(y/3)^2 + (y/3)^3`

                            `= {x^3}/8 + 3{x^2y}/12 + 3{xy^2}/18 + {y^3}/27`

   `c) ({2x}/3-2y)^3 = ({2x}/3)^3 – 3({2x}/3)^2*2y + 3{2x}/3*(2y)^2 – (2y)^3`

                             `= {8x^3}/27 – 2y({4x^2}/3) + 8xy^2 – 8y^3`

   `d) (x+y)^3+(x-y)^3 = x^3 + 3x^2y + 3xy^2 +y^3 + x^3 – 3x^2y + 3xy^3 – y^3`

                                   `= 2x^3 + 6xy^2`

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2. Đáp án:

    

   Giải thích các bước giải:

   $a,\left (\dfrac{x}{2}-y \right )^3$

   $=\dfrac{x^3}{8}-\dfrac{3x^2y}{4}+\dfrac{3xy^2}{2} – y^3$

   $b,\left (\dfrac{x}{2}+\dfrac{y}{3} \right )^3$

   $=\dfrac{x^3}{8}+\dfrac{3x^2y}{12}+\dfrac{3xy^2}{18}+\dfrac{y^3}{27}$

   $c,\left (\dfrac{2x}{3}-2y \right )^3$

   $=\dfrac{8x^3}{27}-\dfrac{24x^2y}{9}+\dfrac{24xy^2}{3}-8y^3$

   $d,(x+y)^3+(x-y)^3$

   $=(x+y+x-y)[(x+y)^2-(x-y)(x+y)+(x-y)^2]$

   $=2x[x^2+2xy+y^2-x^2+y^2+x^2-2xy+y^2]$

   $=2x[(x^2-x^2+x^2)+(2xy-2xy)+(y^2+y^2+y^2)]$

   $=2x[x^2+3y^2]$

   $=2x^3+6xy^2$

   Chúc bạn học tốt.

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