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

\(m\left( {x + y + 1} \right)\).

\(m\left( {x + y + m} \right)\).

\(m\left( {x + y} \right)\).

\(m\left( {x + y - 1} \right)\)

${(5{\rm{x}} - 4)^2} - 49{{\rm{x}}^2} =  - 8\left( {3x + 1} \right)\left( {x + 2} \right)$.                               

\({(5{\rm{x}} - 4)^2} - 49{{\rm{x}}^2} = \left( {3x - 1} \right)\left( {x + 2} \right)\).         

\({(5{\rm{x}} - 4)^2} - 49{{\rm{x}}^2} =  - 8\left( {3x - 1} \right)\left( {x - 2} \right)\).

\({(5{\rm{x}} - 4)^2} - 49{{\rm{x}}^2} =  - 8\left( {3x - 1} \right)\left( {x + 2} \right)\).

${\left( {x - 1} \right)^3} + 2{\left( {x - 1} \right)^2} = {\left( {x - 1} \right)^2}\left( {x + 1} \right)$.

${\left( {x - 1} \right)^3} + 2\left( {x - 1} \right) = \left( {x - 1} \right)\left[ {{{\left( {x - 1} \right)}^2} + 2} \right]$.

${\left( {x - 1} \right)^3} + 2{\left( {x - 1} \right)^2} = \left( {x - 1} \right)\left[ {{{\left( {x - 1} \right)}^2} + 2x - 2} \right]$.

${\left( {x - 1} \right)^3} + 2{\left( {x - 1} \right)^2} $$= \left( {x - 1} \right)\left( {x + 3} \right)$.

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

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

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

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

$x = 2;\,x =  - \dfrac{1}{3}$.

$x =  - 2;\,x = \dfrac{1}{3}$.

$x = 2;\,x = 3$.

\(x = 2;\,x = \dfrac{1}{3}\).

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