Đáp án:

$\begin{array}{l}
1)a)x\left( {x - 3} \right) - 2x + 6 = 0\\
 \Leftrightarrow x\left( {x - 3} \right) - 2\left( {x - 3} \right) = 0\\
 \Leftrightarrow \left( {x - 3} \right)\left( {x - 2} \right) = 0\\
 \Leftrightarrow x = 3/x = 2\\
Vậy\,x = 3;x = 2\\
b)x - 2\sqrt 3 {x^2} + 3{x^3} = 0\\
 \Leftrightarrow x\left( {3{x^2} - 2\sqrt 3 x + 1} \right) = 0\\
 \Leftrightarrow x{\left( {\sqrt 3 x - 1} \right)^2} = 0\\
 \Leftrightarrow x = 0/x = \dfrac{1}{{\sqrt 3 }} = \dfrac{{\sqrt 3 }}{3}\\
Vậy\,x = 0;x = \dfrac{{\sqrt 3 }}{3}\\
c)\left( {x - 1} \right).\left( {{x^2} + x + 1} \right) = {x^2}\left( {x - 9} \right) + 2{x^2} + 6\\
 \Leftrightarrow {x^3} - 1 = {x^3} - 9{x^2} + 2{x^2} + 6\\
 \Leftrightarrow 7{x^2} = 7\\
 \Leftrightarrow {x^2} = 1\\
 \Leftrightarrow x = 1;x =  - 1\\
Vậy\,x = 1;x =  - 1\\
B2)\\
A = 4{a^2}{b^2} + 4ab + 1\\
 = {\left( {2ab} \right)^2} + 2.2ab + 1\\
 = {\left( {2ab + 1} \right)^2} \ge 0
\end{array}$

Vậy giá trị biểu thức không âm với mọi a,b

`a) x(x - 3) - 2x  + 6 = 0`

`⇔x ( x - 3) - 2 ( x - 3) = 0`

`⇔(x - 3)(x - 2) = 0`

⇔ \(\left[ \begin{array}{l}x - 3 = 0\\x - 2 = 0\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x = 3\\x = 2\end{array} \right.\) 

`b)x - 2√3 x^2 + 3x^3 = 0`

`⇔x ( 1- 2√3 x + 3x^2) = 0`

`⇔ x [1^2 - 2 . 1 . √3 x + (√3 x)^2] = 0`

`⇔ x ( 1- √3 x)^2 = 0`

⇔ \(\left[ \begin{array}{l}x = 0\\1- √3 x= 0\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x = 0\\x =\dfrac{ 1}{√3}\end{array} \right.\) 

`c)(x - 1)(x^2 + x + 1) = x^2 (x - 9) + 2x^2 + 6`

`⇔x^3 - 1 = x^3 - 9x^2 + 2x^2 + 6`

`⇔x^3 - 1 - x^3 + 9x^2 - 2x^2 - 6 = 0`

`⇔7x^2 - 7 = 0`

`⇔7(x^2 - 1)=0`

`⇔7(x - 1)(x + 1) = 0`

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

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

`2`

`4a^2b^2 + 4ab + 1`

`=(2ab)^2 + 2 .2ab . 1 + 1^2`

`=(2ab + 1)^2`

`(2ab + 1)^2 ≥ 0 ∀ x ∈ R`

`⇒4a^2b^2 + 4ab + 1 ≥ 0 ∀ x ∈ R`

`⇒4a^2b^2 + 4ab + 1` không âm