1) x3 – 5×2 + 6x = 0 2) 2×3 + 3×2 – 32x = 48 3) (x2 – 2x + 1) – 4 =0 4) 4×2 + 4x + 1 = x2 5) x2 – 5x + 6 = 0 6) x3 + 3×2 + 2

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}\)