Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2,\\
a,\\
\frac{2}{3}{a^2}b.\left( { - 3a{b^2} + 9{a^3}{b^3}} \right)\\
= \frac{2}{3}{a^2}b.\left( { - 3a{b^2}} \right) + \frac{2}{3}{a^2}b.9{a^3}{b^3}\\
= - 2{a^3}{b^3} + 6{a^5}{b^4}\\
b,\\
\left( {2x - 3} \right).\left( {{x^2} + 2x - 4} \right)\\
= 2{x^3} + 4{x^2} - 8x - 3{x^2} - 6x + 12\\
= 2{x^3} + {x^2} - 14x + 12\\
3,\\
a,\\
x.\left( {3x + 2} \right) + {\left( {x + 1} \right)^2} - \left( {2x - 5} \right)\left( {2x + 5} \right) = - 12\\
\Leftrightarrow \left( {3{x^2} + 2x} \right) + \left( {{x^2} + 2x + 1} \right) - \left( {{{\left( {2x} \right)}^2} - {5^2}} \right) = - 12\\
\Leftrightarrow 3{x^2} + 2x + {x^2} + 2x + 1 - 4{x^2} + 25 = - 12\\
\Leftrightarrow 4x + 26 = - 12\\
\Leftrightarrow 4x = - 38\\
\Leftrightarrow x = - \frac{{19}}{2}\\
b,\\
3x - 27x = 0\\
\Leftrightarrow - 24x = 0\\
\Leftrightarrow x = 0
\end{array}\)
