Đáp án:
\(\begin{array}{l}
B1:\\
a)10{x^3} + 6{x^2}\\
b) - 4{x^2} + 8x + 21\\
c)15{x^3} + 10x\\
d) - 6{x^2} + 2x + 10\\
B2:\\
a)3{x^2} + 27\\
b) - 27 - 3{x^2}\\
B3:\\
a)4{x^2}\\
b)4{x^2}\\
B4:\\
a)\left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = - 2
\end{array} \right.\\
b)\left[ \begin{array}{l}
x = \dfrac{2}{3}\\
x = - 2
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)10{x^3} + 6{x^2}\\
b)14x - 4{x^2} + 21 - 6x\\
= - 4{x^2} + 8x + 21\\
c)15{x^3} + 10x\\
d)6x - 6{x^2} + 10 - 4x\\
= - 6{x^2} + 2x + 10\\
B2:\\
a)8{x^3} + 27 - 8{x^3} + 3{x^2}\\
= 3{x^2} + 27\\
b)8{x^3} - 27 - 8{x^3} - 3{x^2}\\
= - 27 - 3{x^2}\\
B3:\\
a){x^2} - 4x + 4 + 2{x^2} - 8 + {x^2} + 4x + 4\\
= 4{x^2}\\
b){x^2} - 6x + 9 + 2{x^2} - 18 + {x^2} + 6x + 9\\
= 4{x^2}\\
B4:\\
a)x\left( {2x - 3} \right) - 2\left( {3 - 2x} \right) = 0\\
\to x\left( {2x - 3} \right) + 2\left( {2x - 3} \right) = 0\\
\to \left( {2x - 3} \right)\left( {x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = - 2
\end{array} \right.\\
b)x\left( {3x - 2} \right) + 2\left( { - 2 + 3x} \right) = 0\\
\to \left( {3x - 2} \right)\left( {x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
3x - 2 = 0\\
x + 2 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{2}{3}\\
x = - 2
\end{array} \right.
\end{array}\)
