x² - 9 = 2(x+3)²

x² - 3³ - 2(x +3 )² = 0 

(x-3)(x+3) -2(x+3)² = 0

(x+3)[(x-3)(x+3) -2 ] = 0

(x+3)(x² -9 -2 ) =0 

(x+3)(x² -11) = 0

TH1 : x+3=0

           x =-3

TH2 : x² -11 =0 

          x² =11 

          x = √11

x² - 9 = 2(x + 3)²

(x² - 9) - 2(x + 3)² = 0

(x - 3)(x + 3) - 2(x + 3)² = 0

(x + 3)[x - 3 - 2(x + 3)] = 0

(x + 3)(x - 3 - 2x - 6) = 0

(x + 3)(-x - 9) = 0

x + 3 = 0 hoặc -x - 9 = 0

*) x + 3 = 0

x = -3

*) -x - 9 = 0

x = -9

Vậy x = -9; x = -3

Lời giải:
a. $5x^2-10xy=5x(x-2y)$

b. $3x(x-y)-6(x-y)=(x-y)(3x-6)=3(x-y)(x-2)$
c. $2x(x-y)-4y(y-x)=2x(x-y)+4y(x-y)=(x-y)(2x+4y)=2(x-y)(x+2y)$

d. $9x^2-9y^2=9(x^2-y^2)=9(x-y)(x+y)$

e. $x^2-xy-x+y=(x^2-xy)-(x-y)=x(x-y)-(x-y)=(x-y)(x-1)$

f. $xy-xz-y+z=(xy-y)-(xz-z)=y(x-1)-z(x-1)=(x-1)(y-z)$

A /B/C/D

Lời giải:
$6x^2-x-2=0$

$\Leftrightarrow x^2-\frac{x}{6}-\frac{1}{3}=0$

$\Leftrightarrow (x^2-\frac{x}{6}+\frac{1}{12^2})-\frac{49}{144}=0$

$\Leftrightarrow (x-\frac{1}{6})^2=\frac{49}{144}$

$\Rightarrow x-\frac{1}{6}=\frac{7}{12}$ hoặc $x-\frac{1}{6}=\frac{-7}{12}$

$\Rightarrow x=\frac{3}{4}$ hoặc $x=\frac{-5}{12}$

a)

\(A=\left(2x\right)^3-1-7x^3-7=8x^3-1-7x^3-7=x^3-8\)

b)

Thay \(x=-\dfrac{1}{2}\) vào biểu thức A:

\(A=\left(-\dfrac{1}{2}\right)^3-8=-\dfrac{65}{8}\)