Đáp án:

$\begin{array}{l}
b){x^2}\left( {x - 5} \right) - 4x + 20 = 0\\
 \Rightarrow {x^2}\left( {x - 5} \right) - 4.\left( {x - 5} \right) = 0\\
 \Rightarrow \left( {x - 5} \right)\left( {{x^2} - 4} \right) = 0\\
 \Rightarrow \left( {x - 5} \right)\left( {x - 2} \right)\left( {x + 2} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 5\\
x = 2\\
x =  - 2
\end{array} \right.\\
Vậy\,x \in \left\{ { - 2;2;5} \right\}\\
c)Dkxd:x \ne 0;x \ne 3\\
\dfrac{{x + 3}}{{x - 3}} - \dfrac{1}{x} = \dfrac{3}{{{x^2} - 3x}}\\
 \Rightarrow \dfrac{{x + 3}}{{x - 3}} - \dfrac{1}{x} = \dfrac{3}{{x\left( {x - 3} \right)}}\\
 \Rightarrow \dfrac{{x\left( {x + 3} \right) - \left( {x - 3} \right)}}{{x\left( {x - 3} \right)}} = \dfrac{3}{{x\left( {x - 3} \right)}}\\
 \Rightarrow {x^2} + 3x - x + 3 = 3\\
 \Rightarrow {x^2} + 2x = 0\\
 \Rightarrow x\left( {x + 2} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 0\left( {ktm} \right)\\
x =  - 2\left( {tm} \right)
\end{array} \right.\\
Vậy\,x =  - 2\\
d)\left( {2{x^2} + 1} \right)\left( {4x - 3} \right) = \left( {2{x^2} + 1} \right)\left( {x - 12} \right)\\
 \Rightarrow \left( {2{x^2} + 1} \right)\left( {4x - 3 - x + 12} \right) = 0\\
 \Rightarrow \left( {2{x^2} + 1} \right)\left( {3x + 9} \right) = 0\\
 \Rightarrow 3x + 9 = 0\left( {do:2{x^2} + 1 > 0} \right)\\
 \Rightarrow x =  - 3\\
Vậy\,x =  - 3\\
e)4{x^2} - 12x + 5 = 0\\
 \Rightarrow 4{x^2} - 10x - 2x + 5 = 0\\
 \Rightarrow 2x\left( {2x - 5} \right) - \left( {2x - 5} \right) = 0\\
 \Rightarrow \left( {2x - 5} \right)\left( {2x - 1} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x = \dfrac{1}{2}
\end{array} \right.\\
g) - {x^2} + 5x - 6 = 0\\
 \Rightarrow {x^2} - 5x + 6 = 0\\
 \Rightarrow {x^2} - 2x - 3x + 6 = 0\\
 \Rightarrow \left( {x - 2} \right)\left( {x - 3} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 2\\
x = 3
\end{array} \right.\\
Vậy\,x = 2;x = 3
\end{array}$

Đáp án:

$b, x²(x - 5) - 4x + 20 = 0 $

$⇔ x²(x - 5) - 4(x - 5) = 0 $

$⇔ (x - 5)(x² - 4) = 0 $

$⇔$ \(\left[ \begin{array}{l}x-5=0\\x^2-4=0\end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}x=5\\x=±2\end{array} \right.\) 

$c, \frac{x+3}{x-3} -\frac{1}{x} =\frac{3}{x^2-3x}$    $(TXĐ : x\neq0 ; x\neq 3) $

$⇔ \frac{(x+3)x-(x-3)}{x(x-3)} =\frac{3}{x(x-3)}$ 

$⇔ x² + 3x - x + 3 = 3 $

$⇔ x² + 2x = 0 $

$⇔ x(x + 2) = 0 $

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

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

$Vậy$ $x = - 2$

$d, (2x² + 1)(4x - 3) = (2x² + 1)(x - 12) $

$⇔(2x² + 1)(4x - 3) - (2x² + 1)(x - 12) = 0$

$⇔ (2x² + 1)(4x - 3 - x + 12)  =0 $

$⇔ (2x² + 1)(3x + 9) =0$

$⇔ 3x + 9 = 0$

$⇔ x = -3 $

$Vậy$ $x = -3$

$e, 4x² - 12x + 5 = 0$

$⇔ 4x² - 2x - 10x + 5 = 0 $

$⇔ 2x(2x - 1) - 5(2x - 1) = 0 $

$⇔ (2x - 1)(2x - 5)  =0 $

$⇔$ \(\left[ \begin{array}{l}x=\frac{1}{2}\\x=\frac{5}{2}\end{array} \right.\) 

$Vậy$ $x =\frac{1}{2} ; x = \frac{5}{2} $

$g,  - x² + 5x - 6 = 0$

$⇔ - x² + 2x + 3x - 6 = 0 $

$⇔ -x(x - 2) + 3(x - 2) = 0$

$⇔ (x - 2)(3 - x) = 0$

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

$Vậy$ $x = 3 ; x = 2$