Question

Помогите пожалуйста с алгеброй 8 класс

Answer 1

$\dfrac{x^2-x}{x^2-6x+9}-\dfrac13=\dfrac{3-x}{3x-9}\\\\\dfrac{x(x-1)}{{(x-3)(x-3)}^{*}}-\dfrac13=\dfrac{-(x-3)}{3(x-3)}\\\\\dfrac{x(x-1)}{(x-3)(x-3)}-\dfrac13=-\dfrac{1}3\\\\\dfrac{x(x-1)}{(x-3)(x-3)}-\dfrac13+\dfrac13=0\\\\\\\dfrac{x(x-1)}{(x-3)(x-3)}=0\qquad ODZ:\ \ (x-3)(x-3)\ne0\ \ <=>\ \ x\ne3$

$x(x-1)=0\\\\\left[\begin{array}{lcl}x_1=0\\x-1=0\end{array}\quad=>\quad\left[\begin{array}{lcl}x_1=0\\x_2=1\end{array}$

Ответ:  $x_1=0;\ x_2=1$

*  $x^2-6x+9=0$

По т. Виета

$\begin{cases}x_1+x_2=-\dfrac{b}{a}\\x_1\cdot x_2 = \dfrac{c}{a}\end{cases}\Longleftrightarrow\quad\begin{cases}x_1+x_2=6\\x_1\cdot x_2 =9\end{cases}\Longleftrightarrow\quad\begin{cases}x_1=3\\x_2=3\end{cases}$

$ax^2+bx+c=a(x-x_1)(x-x_2)$

$x^2-6x+9=(x-3)(x-3)$