a) (5x+1)² - (5x-3).(5x+3) = 0

25x² + 10x + 1 - 25x² + 9 = 0

10x + 10 = 0

10.(x+1) = 0

=> x + 1 = 0 => x = - 1

b) (x+3).(x² - 3x + 9) - x.(x-2).(x+2) = 0

x³ + 27 - x.(x² - 4) = 0

x³ + 27 - x³ + 4x = 0

27 + 4x = 0

4x = - 27

x = -27/4

c) 3x.(x-2) - x + 2= 0

3x.(x-2) - (x-2) = 0

(x-2).(3x-1) = 0

=> x - 2 =0 => x = 2

3x-1 = 0 => 3x = 1 => x = 1/3

d) x.(2x-3) - 2.(3-2x) = 0

x.(2x-3) + 2.(2x-3) = 0

(2x-3).(x+2) = 0

=> 2x - 3 = 0 => 2x =  3 => x = 3/2

x+ 2 = 0 => x = -2

KL:...\

Lời giải:
PT $\Leftrightarrow (2x^2+1)^2-(4x+12)^2+11(2x^2+4x+13)=0$

$\Leftrightarrow (2x^2+1-4x-12)(2x^2+1+4x+12)+11(2x^2+4x+13)=0$

$\Leftrightarrow (2x^2-4x-11)(2x^2+4x+13)+11(2x^2+4x+13)=0$

$\Leftrightarrow (2x^2+4x+13)(2x^2-4x)=0$

\(\Rightarrow \left[\begin{matrix} 2x^2+4x+13=0\\ 2x^2-4x=0\end{matrix}\right.\)

Nếu $2x^2+4x+13=0\Leftrightarrow 2(x+1)^2=-11< 0$ (vô lý)

Nếu $2x^2-4x=0\Leftrightarrow 2x(x-2)=0\Rightarrow x=0$ hoặc $x=2$

\(\left(2x^2+1\right)^2-16\left(x+3\right)^2+11\left(2x^2+4x+13\right)=0\)

...

\(4x^4+10x^2-52x=0\)

\(2x\left(2x^3+5x-26\right)=0\)

\(2x\left(2x^2+4x+13\right)\left(x-2\right)=0\)

\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

Tự tính tiếp vs : \(2x^2+4x+13=0\)

a) x² - x = 0 <=> x(x - 1) = 0 <=> x = 0 hoặc x - 1 = 0 <=> x = 0 hoặc x = 1

Vậy : S = {0; 1}.

b) x² - 2x = 0 <=> x(x - 2) <=> x = 0 hoặc x - 2 = 0 <=> x = 0 hoặc x = 2

Vậy : S = {0; 2).

(Bài này dễ mà)

`a)(x-1)^2-(x-2)(x+2)`

`=x^2-2x+1-(x^2-4)`

`=-2x+5`

`b)(2x+4)(8x-3)(4x+1)^2`

`=(16x^2-6x+32x-12)(16x^2+8x+1)`

`=(16x^2-26x-12)(16x^2+8x+1)`

`=256x^4+128x^3+16x^2-416x^3-208x^2-26x-192x^2-96x-12`

`=256x^4-288x^3-384x^2-122x-12`

`c)(a+2)^3-a(a-3)^2`

`=a^3+6a^2+12a+8-a(a^2-6a+9)`

`=a^3+6a^2+12a+8-a^3+6a^2-9a`

`=12a^2+3a+8`

a) \(x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)

b) \(4x^2+2x+2=0\)

thank

mình vt thiếu dấu bằng ở trc (x-3)