Đáp án:
12) \(\left[ \begin{array}{l}
x = 0\\
x = - \dfrac{5}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1){x^3} - 5{x^2} + 6x = 0\\
\to x\left( {{x^2} - 5x + 6} \right) = 0\\
\to x\left( {x - 2} \right)\left( {x - 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 2\\
x = 3
\end{array} \right.\\
2)2{x^3} + 3{x^2} - 32x = 48\\
\to 2{x^3} - 8{x^2} + 11{x^2} - 44x + 12x - 48 = 0\\
\to 2{x^2}\left( {x - 4} \right) + 11x\left( {x - 4} \right) + 12\left( {x - 4} \right) = 0\\
\to \left( {x - 4} \right)\left( {2{x^2} + 11x + 12} \right) = 0\\
\to \left( {x - 4} \right)\left( {2x + 3} \right)\left( {x + 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = 4\\
x = - \dfrac{3}{2}\\
x = - 4
\end{array} \right.\\
3)\left( {{x^2} - 2x + 1} \right) - 4 = 0\\
\to {\left( {x - 1} \right)^2} = 4\\
\to \left| {x - 1} \right| = 2\\
\to \left[ \begin{array}{l}
x - 1 = 2\\
x - 1 = - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.\\
4)4{x^2} + 4x + 1 = {x^2}\\
\to {\left( {2x + 1} \right)^2} = {x^2}\\
\to \left| {2x + 1} \right| = \left| x \right|\\
\to \left[ \begin{array}{l}
2x + 1 = x\left( {DK:x \ge 0} \right)\\
2x + 1 = - x\left( {DK:x < 0} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\left( l \right)\\
x = - \dfrac{1}{3}\left( {TM} \right)
\end{array} \right.\\
5){x^2} - 5x + 6 = 0\\
\to \left( {x - 2} \right)\left( {x - 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 3
\end{array} \right.\\
6){x^3} + 3{x^2} + 2x = 0\\
\to x\left( {{x^2} + 3x + 2} \right) = 0\\
\to x\left( {x + 1} \right)\left( {x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = - 1\\
x = - 2
\end{array} \right.\\
8){x^2} - 2x + 1 = 25\\
\to {\left( {x - 1} \right)^2} = 25\\
\to \left| {x - 1} \right| = 5\\
\to \left[ \begin{array}{l}
x - 1 = 5\\
x - 1 = - 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 6\\
x = - 4
\end{array} \right.\\
9){x^2} - x = 0\\
\to x\left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
10){x^2} - 2x = 0\\
\to x\left( {x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.\\
11){x^2} - 3x = 0\\
\to x\left( {x - 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 3
\end{array} \right.\\
12)\left( {x + 1} \right)\left( {x + 4} \right) = \left( {2 - x} \right)\left( {x + 2} \right)\\
\to {x^2} + 5x + 4 = 4 - {x^2}\\
\to 2{x^2} + 5x = 0\\
\to x\left( {2x + 5} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = - \dfrac{5}{2}
\end{array} \right.
\end{array}\)
