Tìm x, biết: a) 5x(x – 2000) – x + 2000 = 0 b) x3 – 13x = 0

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

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

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

`<=>` \(\left[ \begin{array}{l}x-200=0\\5x-1=0\end{array} \right.\)

`<=>` \(\left[ \begin{array}{l}x=2000\\x=\dfrac{1}{5}\end{array} \right.\) 

 Vậy `x=2000; x=1/5`

`b) x^3-13x=0`

`<=> x(x^2-13)=0`

`<=>` \(\left[ \begin{array}{l}x=0\\x^2-13=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=0\\x^2=13\end{array} \right.\) \(\left[ \begin{array}{l}x=0\\\left[\begin{matrix} x=\sqrt{13}\\ x=\sqrt{-13}\end{matrix}\right.\end{array} \right.\) 

Vậy `x=0;x=\sqrt{13}; \sqrt{-13}`

Đáp án:

1. a) $S=\left\{\dfrac{1}{5};2000\right\}$
2. b) $S=\{0;±\sqrt{13}\}$

Giải thích các bước giải:

1. a) $5x(x-2000)-x+2000=0$

$⇔5x(x-2000)-(x-2000)=0$

$⇔(x-2000)(5x-1)=0$

$⇔\left[ \begin{array}{l}x-2000=0\\5x-1=0\end{array} \right.⇔\left[ \begin{array}{l}x=2000\\x=\dfrac{1}{5}\end{array} \right.$

Vậy $S=\left\{\dfrac{1}{5};2000\right\}$.

1. b) $x^3-13x=0$

$⇔x(x^2-13)=0$

$⇔x(x-\sqrt{13})(x+\sqrt{14})=0$

$⇔\left[ \begin{array}{l}x=0\\x-\sqrt{13}=0\\x+\sqrt{13}\end{array} \right.⇔\left[ \begin{array}{l}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{array} \right.$

Vậy $S=\{0;±\sqrt{13}\}$.