a, 3x ³ - 3x = 0

=> 3x ( x ² - 1 ) = 0

=> \(\orbr{\begin{cases}3x=0\\x^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x^2=1\end{cases}\Rightarrow[}\begin{cases}x=0\\x=1\\x=-1\end{cases}}\)

b, x ( x - 2 ) + ( x - 2 ) = 0

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

=> \(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)

c, 5x ( x - 2000 ) - x + 2000 = 0

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

=> \(\orbr{\begin{cases}x-2000=0\\5x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2000\\x=\frac{1}{5}\end{cases}}}\)

Bài giải:

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

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

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

Hoặc 5x - 1 = 0 => 5x = 1 => x = $\frac{1}{5}$

Vậy x = $\frac{1}{5}$; x = 2000

b) x³ – 13x = 0

x(x² - 13) = 0

Hoặc x = 0

Hoặc x² - 13 = 0 => x² = 13 => x = ±√13

Vậy x = 0; x = ±√13

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

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

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

\(\Leftrightarrow\) x-2000=0 hoặc 5x-1=0

\(\Leftrightarrow\) x=2000 hoặc x=\(\dfrac{1}{5}\)

b) \(x^3-13x=0\)

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

\(\Leftrightarrow x=0\) hoặc \(x^2-13=0\)

\(\Leftrightarrow x=0\) hoặc \(x=13\) hoặc \(x=-13\)

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

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

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

Hoặc 5x - 1 = 0 => 5x = 1 => x =

Vậy x = ; x = 2000

b) x³ – 13x = 0

x(x² - 13) = 0

Hoặc x = 0

Hoặc x² - 13 = 0 => x² = 13 => x = ±√13

Vậy x = 0; x = ±√13

a ) \(5x\left(x-2000\right)-x+2000=0\)

\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)

\(\Leftrightarrow\left(x-2000\right)\left(5x-1\right)=0\)

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

b ) \(x^3-13x=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x^2-13=0\Rightarrow\left[\begin{array}{nghiempt}x=\sqrt{13}\\x=-\sqrt{13}\end{array}\right.\end{array}\right.\)

a) \(3x^3-3x=0\)

\(\Rightarrow3x\left(x^2-1\right)=0\)

\(\Rightarrow3x\left(x-1\right)\left(x+1\right)=0\)

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

Vậy \(x\in\left\{0;\pm1\right\}\)

b) \(x\left(x-2\right)+x-2=0\)

\(\Rightarrow x\left(x-2\right)+\left(x-2\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

Vậy \(x\in\left\{-1;2\right\}\)

c) \(5x\left(x-2000\right)-x+2000=0\)

\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)

\(\Rightarrow\left(5x-1\right)\left(x-2000\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\x-2000=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2000\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{1}{5};2000\right\}\)

a)    5x(x - 2000) - (x - 2000) = 0

tương đương (x - 2000)(5x - 1) = 0

tương đương x = 2000 hoặc x = 1/5

b)   x(x^2 -13) = 0

\(x\left(x-\sqrt{13}\right)\left(x+\sqrt{13}\right)=0\)

tương đương x = 0 hoặ x = \(\sqrt{13}\)hoặc x = \(-\sqrt{13}\)

=> 5.x(x - 2000) - (x - 2000) = 0

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

=> x - 2000 = 0 => x = 2000

hoặc 5x - 1 = 0 => 5x = 1 => x = 1/5

Vậy x = 2000; x = 1/5

Supper Saydan Goku

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

<=>5x(x-2000)-(x-2000)=0

<=>(5x-1)(x-2000)=0

<=>5x-1=0 hoặc x-2000=0

<=>5x=1 hoặc x=2000

5x=1,Mà x>1 =>loại

=>x=2000

Ta có:5\(\times\)(x-2000)-x+2000=0

         x\(\times\)5-2000\(\times\)5-x+2000=0

         x\(\times\)4-8000=0

        \(\Rightarrow\)x\(\times\)4=8000

             x=8000\(\div\)4=2000

Vậy x bằng 2000.