Đáp án:
B1:
a) \(3{x^3} + 2{x^2} - 8x\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)\left( {x + 2} \right)\left( {3{x^2} - 4x} \right)\\
= 3{x^3} - 4{x^2} + 6{x^2} - 8x\\
= 3{x^3} + 2{x^2} - 8x\\
b){x^2} - 4x + 3 + 2{x^2} - 7x - 4 - 3{x^2} + 2x - 5\\
= - 9x - 6\\
B2:\\
a{\left( {2x + 1} \right)^2} - 2\left( {2x + 1} \right)\left( {2x + 3} \right) + {\left( {2x + 3} \right)^2}\\
= {\left( {2x + 1 - 2x - 3} \right)^2} = 4\\
b){\left( {x - 3y} \right)^2} - 25{z^2}\\
= \left( {x - 3y - 5z} \right)\left( {x - 3y + 5z} \right)\\
c){x^2} - 25x + 14x\\
= {x^2} - x = x\left( {x - 1} \right)\\
B3:\\
\left( {x - 3} \right)\left( {x + 2} \right) - \left( {x - 3} \right)\left( {2x + 7} \right) = 0\\
\to \left( {x - 3} \right)\left( {x + 2 - 2x - 7} \right) = 0\\
\to \left[ \begin{array}{l}
x - 3 = 0\\
- x - 5 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 5
\end{array} \right.
\end{array}\)
