a.

`4x(x + 1) = 8(x + 1)`

`⇔4x(x + 1) - 8(x + 1) = 0`

`⇔(x + 1)(4x - 8) = 0`

`⇔4(x + 1)(x - 2) = 0`

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

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

Vậy `S = {-1; 2}`

b.

`x(x - 1) - 2(1 - x) = 0`

`⇔x(x - 1) + 2(x - 1) = 0`

`⇔(x - 1)(x + 2) = 0`

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

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

Vậy `S = {-2; 1}`

c.

`2x(x - 2) - (2 - x)^2 = 0`

`⇔-2x(2 - x) - (2 - x)^2 = 0`

`⇔(2 - x)(-2x - 2 + x) = 0`

`⇔(2 - x)(-x - 2) = 0`

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

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

Vậy `S = {-2; 2}`

d.

`(x - 3)^3 + (3 - x) = 0`

`⇔(x - 3)^3 - (x - 3) = 0`

`⇔(x - 3)[(x - 3)^2 - 1] = 0`

`⇔(x - 3)(x - 3 - 1)(x - 3 + 1) = 0`

`⇔(x - 3)(x - 4)(x - 2) = 0`

`⇔`\(\left[ \begin{array}{l}x-3=0\\x-4=0\\x-2=0\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=3\\x=4\\x=2\end{array} \right.\) 

Vậy `S = {2; 3; 4}`

e.

`5x(x - 2) - (2 - x) = 0`

`⇔5x(x - 2) + (x - 2) = 0`

`⇔(x - 2)(5x -1 ) = 0`

`⇔`\(\left[ \begin{array}{l}x-2=0\\5x-1=0\end{array} \right.\) 

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

Vậy `S = {1/5; 2}`

g.

`5x(x - 2000) - x + 2000 = 0`

`⇔5x(x - 2000) - (x - 2000) = 0`

`⇔(x - 2000)(5x - 1) = 0`

`⇔`\(\left[ \begin{array}{l}x-2000=0\\5x-1=0\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=2000\\x=\dfrac{1}{5}\end{array} \right.\) 

Vậy `S = {1/5; 2000}`

h.

`x^2 - 4x = 0`

`⇔x(x - 4) = 0`

`⇔`\(\left[ \begin{array}{l}x=0\\x-4=0\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=0\\x=4\end{array} \right.\) 

Vậy `S = {0; 4}`

k.

`(1 - x)^2 - 1 + x = 0`

`⇔(1 - x)^2 - (1 - x) = 0`

`⇔(1 - x)(1 - x - 1) = 0`

`⇔-x(1 - x) = 0`

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

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

Vậy `S = {0; 1}`

m.

`x + 6x^2 = 0`

`⇔x(1 + 6x) = 0`

`⇔`\(\left[ \begin{array}{l}x=0\\1+6x=0\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=0\\x=\dfrac{-1}{6}\end{array} \right.\) 

Vậy `S = {-1/6; 0}`

n.

`(x + 1) = (x + 1)^2`

`⇔(x + 1) - (x + 1)^2 = 0`

`⇔(x + 1)(1 - x - 1) = 0`

`⇔-x(x + 1) = 0`

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

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

Vậy `S = {-1; 0}`

$a, 4x ( x + 1 ) = 8 ( x + 1 )$

$(=) 4x ( x + 1 ) -8 ( x + 1 )=0$

$(=)(4x-8)(x+1)=0$

$(=)\left[ \begin{array}{l}4x-8=0\\x+1=0\end{array} \right. (=)\left[ \begin{array}{l}4x=8\\x=-1\end{array} \right. (=)\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.$

$S={2;-1}$

$b, x( x - 1 ) - 2 ( 1 - x ) = 0$

$(=)x(x-1)+2(x-1)=0$
$(=)(x+2)(x-1)=0$

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

$S={-2;1}$

$c, 2x ( x - 2 ) - ( 2 - x )² = 0$

$(=)2x(x-2)-(x-2)²=0$

$(=)(x-2)(2x-x+2)=0$

$(=)(x-2)(x+2)=0$

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

$S={-2;2}$

$d, ( x - 3 )³ + ( 3 - x ) = 0$

$(=)( x - 3 )³-(x-3)=0$

$(=)(x-3)[(x-3)²-1]=0$

$(=)(x-3)(x-3-1)(x-3+1)=0$

$(=)(x-3)(x-4)(x-2)=0$

$\text{TH1: x-3-0(=)x=3}$

$\text{TH2: x-4-0(=)x=4}$

$\text{TH3: x-2-0(=)x=2}$

$S={3;2;4}$

$e, 5x ( x - 2 ) - ( 2 - x ) = 0$

$(=)5x ( x - 2 ) + ( x-2 ) = 0$

$(=)(5x+1)(x-2)=0$

$(=)\left[ \begin{array}{l}x-2=0\\5x+1=0\end{array} \right. (=)\left[ \begin{array}{l}x=2\\5x=-1\end{array} \right.(=)\left[ \begin{array}{l}x=2\\x=\frac{-1}{5}\end{array} \right.$

$S={2;\frac{-1}{5}}$

$g, 5x ( x - 2000 ) - x + 2000 = 0$

$(=)5x(x-2000)-(x-2000)=0$

$(=)(5x-1)(x-2000)=0$

$(=)\left[ \begin{array}{l}x-2000=0\\5x-1=0\end{array} \right. (=)\left[ \begin{array}{l}x=2000\\5x=1\end{array} \right.(=)\left[ \begin{array}{l}x=2000\\x=\frac{1}{5}\end{array} \right.$

$S={2000;\frac{1}{5}}$

$h, x² - 4x = 0$

$(=)x(x-4)=0$

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

$S={0;4}$

$k, ( 1 - x )² - 1 + x = 0$

$(=) ( 1 - x )² -( 1 -x) = 0$

$(=)(1-x)(1-x-1)=0$

$(=)-x(1-x)=0$

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

$S={0;1}$

$m, x + 6x² = 0$

$(=)x(x+6)=0$

$(=)\left[ \begin{array}{l}x=0\\x+6=0\end{array} \right. (=)\left[ \begin{array}{l}x=0\\x=-6\end{array} \right.$

$S={0;-6}$

$n, ( x + 1 ) = ( x + 1 )²$

$(=) ( x + 1 )²=( x + 1 ) $

$(=)( x + 1 )²-( x + 1 )=0$

$(=)(x+1)(x-1-1)=0$

$(=)x(x+1)=0$

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

$S={0;-1}$