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

<=>(x-3)3(x-3)=0

=>x-3=0=>x=3

Không có giá trị x nào thỏa mãn nha bạn

3.(3x-1/2)^3+1/9=0

3.(3x-1/2)^3=0-1/9

3.(3x-1/2)^3=-1/9

(3x-1/2)^3=(-1/9):3

(3x-1/2)^3=-1/27

(3x-1/2)^3=(-1/3)^3

=>3x-1/2=-1/3

3x=(-1/3)+1/2

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

3x=1/6

x=1/6:3

x=1/6.1/3

x=1/18

Vậy x =1/18

a) 4x(x - 5) - (x - 1)(4x - 3) = 5

4x² - 20x - (4x² - 3x - 4x + 3) = 5

4x² - 20x - 4x² + 3x + 4x - 3 = 5

-13x - 3 = 5

\(\Rightarrow\) -13x = 8

\(\Rightarrow\) x = \(\dfrac{-8}{13}\)

b) (3x - 4)(x - 2) = 3x(x - 9) - 3

3x² - 6x - 4x + 8 = 3x² - 27x - 3

3x² - 10x + 8 - 3x² + 27x + 3 = 0

17x + 11 = 0

\(\Rightarrow\) 17x = -11

\(\Rightarrow\) x = \(\dfrac{-11}{17}\)

c) x² - 81 = 0

\(\Rightarrow\) x² = 81

\(\Rightarrow\) x = \(\pm\) 9

d) 3x² - 75 = 0

3(x² - 25) = 0

\(\Rightarrow\) x² - 25 = 0

\(\Rightarrow\) x² = 25

\(\Rightarrow\) x = \(\pm\)5

e) x² - 4x + 3 = 0

x² - x - 3x + 3 = 0

(x² - x) - (3x - 3) = 0

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

xin lỗi vì chữa đề

a) x³ - 3x² + 3x - 1 = 0

<=>(x-1)³=0

<=>x-1=0

<=>x=1

b) x³+6x² + 12x+8 =0

<=>(x+2)³=0

<=>x+2=0

<=>x=-2

c) (x-2)³+6(x+1)²-x³+9=0

<=>x³-6x²+12x-8+6x²+12x+6-x³+9=0

<=>24x+7=0

<=>24x=-7

<=>x=-7/24