Đáp án:
$1)-27\\
2)
7$
Giải thích các bước giải:
$1) 3(x-1)(x^2+x+1)+(x-1)^3-4x(x+1)(x-1)\\
=3(x^3-1)+x^3-3x^2+3x-1-4x(x^2-1)\\
=3x^3-3+x^3-3x^2+3x-1-4x^3+4x\\
=(3x^3+x^3-4x^3)+(-3-1)-3x^2+(3x+4x)\\
=-4-3x^2+7x\\
\Rightarrow -4-3.(-2)^2+7.(-2)=-4-3.3-14=-27\\
2)
(3xy-2)(9x^2y^2+6xy+4)-3xy(3xy+1)^2\\
=(3xy)^3-2^3-3xy(9x^2y^2-6xy+1)\\
=27x^3y^3-8-27x^3y^3+18x^2y^2-3xy\\
=18x^2y^2-3xy-8\\
\Rightarrow 18.(-2010)^2.\left ( \dfrac{-1}{2010} \right )^2-3.(-2010).\left ( \dfrac{-1}{2010} \right )-8\\
=18.2010^2.\dfrac{1}{2010^2}-3.(-2010).\dfrac{1}{-2010}-8\\
=18-3-8\\
=7$
