`a)`

`3x^2 - 5x + 2 = 0`

`⇔ 3x^2 - 2x - 3x + 2 = 0`

`⇔ x(3x-2)-(3x-2)=0`

`⇔ (3x-2)(x-1) = 0`

Trường hợp 1 : `3x - 2 = 0 ⇔ 3x = 2 ⇔ x = 2/3`

Trường hợp 2 : `x - 1 = 0 ⇔ x = 1`

Vậy `S = {1,2/3}`

`b)`

`(2x-3)^2 = x^2 - 10x + 25`

`⇔ 4x^2 - 12x + 9 = x^2 - 10x + 25`

`⇔ 4x^2 - 12x + 9 - x^2 + 10x - 25 = 0`

`⇔ 3x^2 - 2x - 16 = 0`

`⇔ 3x^2 + 6x - 8x - 16 = 0`

`⇔ 3x(x+2) - 8(x+2) = 0`

`⇔ (x+2)(3x-8) = 0`

Trường hợp 1 : `x + 2 = 0 ⇔ x = -2`

Trường hợp 2 : `3x - 8 = 0 ⇔ 3x = 8 ⇔ x = 8/3`

Vậy `S = {2,8/3}`

`c)`

`x^3 - 9x^2 = 0`

`⇔ x^2(x-9) = 0`

Trường hợp 1 : `x^2 = 0 ⇔ |x| = 0 ⇔ x = 0`

Trường hợp 2 : `x - 9 = 0 ⇔ x - 9`

Vậy `S = {0,9}

`d)`

`x^3+4x^2+4x=0`

`⇔ x(x^2+4x+4) = 0`

`⇔ x(x^2+2x*2 + 2^2) = 0`

`⇔ x(x+2)^2 = 0`

Trường hợp 1 : `x = 0`

Trường hợp : `(x+2)^2 = 0 ⇔ x + 2 = 0 ⇔ x = -2`

Vậy `S = {-2,0}`

Đáp án:

Giải thích các bước giải:

$\text{a,3x²-5x+2=0}$

$\text{⇔3x²-3x-2x+2=0}$

$\text{⇔3x(x-1)-2(x-1)=0}$

$\text{⇔(3x-2)(x-1)=0}$

$\text{⇔\(\left[ \begin{array}{l}3x-2=0\\x-1=0\end{array} \right.\) }$

$\text{⇔\(\left[ \begin{array}{l}x=2/3\\x=1\end{array} \right.\) }$

$\text{Vậy S={2/3; 1}}$

$\text{b,(2x-3)²=x²-10x+25}$

$\text{⇔4x²-12x+9=x²-10x+25}$

$\text{⇔4x²-12x+9-x²+10x-25=0}$

$\text{⇔3x²+6x-8x-16=0}$

$\text{⇔3x(x+2)-8(x+2)=0}$

$\text{⇔(3x-8)(x+2)=0}$

$\text{⇔\(\left[ \begin{array}{l}3x-8=0\\x+2=0\end{array} \right.\) }$

$\text{⇔\(\left[ \begin{array}{l}x=8/3\\x=-2\end{array} \right.\) }$

$\text{Vậy S= { 8/3 ; -2 }}$

$\text{c, x³-9x²=0}$

$\text{⇔x²(x-9)=0}$

$\text{⇔\(\left[ \begin{array}{l}x²=0\\x-9=0\end{array} \right.\) }$

$\text{⇔\(\left[ \begin{array}{l}x=0\\x=9\end{array} \right.\) }$

$\text{Vậy S={0; 9} }$

$\text{d, x³+4x²+4x=0}$

$\text{⇔x³+2x²+2x²+4x=0}$

$\text{⇔x²(x+2)+2x(x+2)=0}$

$\text{⇔(x²+2x)(x+2)=0}$

$\text{⇔x(x+2)(x+2)=0}$

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

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

$\text{Vậy S= {0; -2}}$