Cho $27{x^3} - 0,001 = \left( {3x - 0,1} \right)\left( {...} \right)$ . Biểu thức thích hợp điền vào dấu $...$ là

${\left( {x + 2y} \right)^3}$.

${\left( {2x + y} \right)^3}$.

${\left( {2x - y} \right)^3}$.

${\left( {8x + y} \right)^3}$.

${\left( {3x - 2y} \right)^2} - {\left( {2x - 3y} \right)^2} = 5\left( {x - y} \right)\left( {x + y} \right)$.

${\left( {3x - 2y} \right)^2} - {\left( {2x - 3y} \right)^2} = \left( {5x - y} \right)\left( {x - 5y} \right)$.

${\left( {3x - 2y} \right)^2} - {\left( {2x - 3y} \right)^2} = \left( {x - y} \right)\left( {x + y} \right)$.

${\left( {3x - 2y} \right)^2} - {\left( {2x - 3y} \right)^2} = 5\left( {x - y} \right)\left( {x - 5y} \right)$.

${x^2} - 6x + 9 = {\left( {x - 3} \right)^2}$.       

$4{x^2} - 4xy + {y^2} = {\left( {2x - y} \right)^2}$.

${x^2} + x + \dfrac{1}{4} = {\left( {x + \dfrac{1}{2}} \right)^2}$.        

$- {x^2} - 2xy - {y^2} =  - {\left( {x - y} \right)^2}$.

${\left( {xy + 2} \right)^3}$.

${\left( {xy + 8} \right)^3}$.

${x^3}{y^3} + 8$.

${\left( {{x^3}{y^3} + 2} \right)^3}$.

${\left( {a - 3} \right)^2}{\left( {a + 3} \right)^2}$.

${\left( {a + 3} \right)^4}$.

$\left( {{a^2} + 36a + 9} \right)\left( {{a^2} - 36a + 9} \right)$.

${\left( {{a^2} + 9} \right)^2}$

$\left( {\dfrac{{{x^2}}}{4} + 5y} \right)\left( {\dfrac{{{x^4}}}{4} - \dfrac{5}{4}{x^2}y + 5{y^2}} \right)$.

$\left( {\dfrac{{{x^2}}}{4} - 5y} \right)\left( {\dfrac{{{x^4}}}{{16}} + \dfrac{5}{4}{x^2}y + 25{y^2}} \right)$.

$\left( {\dfrac{{{x^2}}}{4} + 5y} \right)\left( {\dfrac{{{x^4}}}{{16}} - \dfrac{5}{4}{x^2}y + 25{y^2}} \right)$.

$\left( {\dfrac{{{x^2}}}{4} + 5y} \right)\left( {\dfrac{{{x^4}}}{{16}} - \dfrac{5}{2}{x^2}y + 25{y^2}} \right)$.