a, A = x^3 - 3x^2 - 3x +1

⇔ A = (x^3 + 1) - (3x^2 + 3x)

⇔ A = (x+1)(x^2 - x + 1)  - 3x(x+1)

⇔A = (x+1)(x^2-x+1-3x)

⇔ A =(x+1)(x^2 - 4x +1)

b, B = x^3 + x^2-36

⇔ B = x^3 - 3x^2 + 4x^2 - 12x + 12x - 36

⇔ B = x^2(x-3) + 4x(x-3) + 12(x-3) 

⇔ B = (x-3)(x^2 + 4x + 12)

* Tìm x

a, x^3 - 9x^2 + 14x = 0

⇔ x^3 - 7x^2 - 2x^2 + 14 = 0

⇔ x(x-2)(x-7) = 0

⇔ x = 0 hoặc x = 2 hoặc x =7

b, x^3 - 5x^2+8x-4=0

⇔ x^3 - 2x^2 - 3x^2 + 6x +2x - 4 = 0

⇔ (x-2)^2(x-1) = 0

⇔\(\left[ \begin{array}{l}(x-2)^2=0\\x-1=0\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=2\\x=1\end{array} \right.\) 

c, 2x^3 + x^2 - 4x -2 = 0

⇔ (2x +1)(x^2-2) = 0

⇔ \(\left[ \begin{array}{l}2x+1=0\\x^2-2=0\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=-1/2\\x=x=√2\end{array} \right.\) 

Đáp án:

2) d. \(\left[ \begin{array}{l}
x =  - \dfrac{1}{2}\\
x = \sqrt 2 \\
x =  - \sqrt 2 
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
1)a.A = {x^3} + {x^2} - 4{x^2} - 4x + x + 1\\
 = {x^2}\left( {x + 1} \right) - 4x\left( {x + 1} \right) + \left( {x + 1} \right)\\
 = \left( {x + 1} \right)\left( {{x^2} - 4x + 1} \right)\\
b.B = {x^3} + {x^2} - 36\\
 = {x^3} - 3{x^2} + 4{x^2} - 12x + 12x - 36\\
 = {x^2}\left( {x - 3} \right) + 4x\left( {x - 3} \right) + 12\left( {x - 3} \right)\\
 = \left( {x - 3} \right)\left( {{x^2} + 4x + 12} \right)\\
2)a.{x^3} - 9{x^2} + 14x = 0\\
 \to x\left( {{x^2} - 9x + 14} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 0\\
{x^2} - 7x - 2x + 14 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 0\\
x\left( {x - 7} \right) - 2\left( {x - 7} \right) = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 0\\
x - 7 = 0\\
x - 2 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 0\\
x = 7\\
x = 2
\end{array} \right.\\
b.{x^3} - 5{x^2} + 8x - 4 = 0\\
 \to {x^3} - {x^2} - 4{x^2} + 4x + 4x - 4 = 0\\
 \to {x^2}\left( {x - 1} \right) - 4x\left( {x - 1} \right) + 4\left( {x - 1} \right) = 0\\
 \to \left[ \begin{array}{l}
x - 1 = 0\\
{x^2} - 4x + 4 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 1\\
{\left( {x - 2} \right)^2} = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 1\\
x = 2
\end{array} \right.\\
c.{x^4} - 2{x^3} + {x^2} = 10\\
 \to {x^2}\left( {{x^2} - 2x + 1} \right) = 10\\
 \to {x^2}{\left( {x - 1} \right)^2} = 10\\
 \to \left[ \begin{array}{l}
x\left( {x - 1} \right) = \sqrt {10} \\
x\left( {x - 1} \right) =  - \sqrt {10} 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
{x^2} - x - \sqrt {10}  = 0\\
{x^2} - x + \sqrt {10}  = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
{x^2} - 2x.\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{4} - \sqrt {10}  = 0\\
{x^2} - 2x.\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{4} + \sqrt {10}  = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
{\left( {x - \dfrac{1}{2}} \right)^2} = \dfrac{1}{4} + \sqrt {10} \\
{\left( {x - \dfrac{1}{2}} \right)^2} = \dfrac{1}{4} - \sqrt {10} \left( l \right)\left( {Do:{{\left( {x - \dfrac{1}{2}} \right)}^2} \ge \forall x} \right)
\end{array} \right.\\
 \to {\left( {x - \dfrac{1}{2}} \right)^2} = \dfrac{1}{4} + \sqrt {10} \\
 \to \left[ \begin{array}{l}
x - \dfrac{1}{2} = \sqrt {\dfrac{1}{4} + \sqrt {10} } \\
x - \dfrac{1}{2} =  - \sqrt {\dfrac{1}{4} + \sqrt {10} } 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{1}{2} + \sqrt {\dfrac{1}{4} + \sqrt {10} } \\
x = \dfrac{1}{2} - \sqrt {\dfrac{1}{4} + \sqrt {10} } 
\end{array} \right.\\
d.2{x^3} + {x^2} - 4x - 2 = 0\\
 \to {x^2}\left( {2x + 1} \right) - 2\left( {2x + 1} \right) = 0\\
 \to \left[ \begin{array}{l}
2x + 1 = 0\\
{x^2} - 2 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - \dfrac{1}{2}\\
x = \sqrt 2 \\
x =  - \sqrt 2 
\end{array} \right.
\end{array}\)