tìm x biết a) 13x(x-2019)-x+2019=0 9x^2-9x=0 04/12/2021 Bởi Maya tìm x biết a) 13x(x-2019)-x+2019=0 9x^2-9x=0
Đáp án: a, $13x(x-2019)-x+2019=0$ $\to 13x(x-2019)-(x-2019)=0$ $\to (x-2019)(13x-1)=0$ $\to \left[\begin{array}{l}x-2019=0\\13x-1=0\end{array}\right.$ $\to \left[\begin{array}{l}x=2019\\x=\dfrac{1}{13}\end{array}\right.$ b, $9x^2-9x=0$ $\to 9x(x-1)=0$ $\to \left[\begin{array}{l}9x=0\\x-1=0\end{array}\right.$ $\to \left[\begin{array}{l}x=0\\x=1\end{array}\right.$ Bình luận
$13x(x-2019)-x+2019=0$ $⇔13x(x-2019)-(x-2019)=0$ $⇔(13x-1)(x-2019)=0$ $⇔\left[ \begin{array}{1}13x=1\\x=2019\end{array} \right.$ $⇔\left[ \begin{array}{1}x=\dfrac{1}{13}\\x=2019\end{array} \right.$ $9x^2-9x=0$ $⇔9x(x-1)=0$ $⇔\left[ \begin{array}{1}x=0\\x=1\end{array} \right.$ Bình luận
Đáp án:
a,
$13x(x-2019)-x+2019=0$
$\to 13x(x-2019)-(x-2019)=0$
$\to (x-2019)(13x-1)=0$
$\to \left[\begin{array}{l}x-2019=0\\13x-1=0\end{array}\right.$
$\to \left[\begin{array}{l}x=2019\\x=\dfrac{1}{13}\end{array}\right.$
b,
$9x^2-9x=0$
$\to 9x(x-1)=0$
$\to \left[\begin{array}{l}9x=0\\x-1=0\end{array}\right.$
$\to \left[\begin{array}{l}x=0\\x=1\end{array}\right.$
$13x(x-2019)-x+2019=0$
$⇔13x(x-2019)-(x-2019)=0$
$⇔(13x-1)(x-2019)=0$
$⇔\left[ \begin{array}{1}13x=1\\x=2019\end{array} \right.$
$⇔\left[ \begin{array}{1}x=\dfrac{1}{13}\\x=2019\end{array} \right.$
$9x^2-9x=0$
$⇔9x(x-1)=0$
$⇔\left[ \begin{array}{1}x=0\\x=1\end{array} \right.$