Question

. решите с ОДЗ, спасибо за ранее​

Answer 1

Ответ:

$\dfrac{9x^2-4}{x-1}=\dfrac{5-10x}{1-x}\\\\\\a)\ \ ODZ:\ \left\{\begin{array}{l}x-1\ne 0\\1-x\ne 0\end{array}\right \ \ \ \Rightarrow \ \ \ x\ne 1\\\\\\b)\ \ \dfrac{9x^2-4}{x-1}-\dfrac{5-10x}{1-x}=0\ \ \to \ \ \ \dfrac{9x^2-4}{x-1}-\dfrac{10x-5}{x-1}=0\ \ ,\\\\\\ \dfrac{9x^2-4-10x+5}{x-1}=0\ \ ,\ \ \ \dfrac{9x^2-10x+1}{x-1}=0\ \ ,\\\\\\9x^2-10x+1=0\ \ ,\\\\\\c)\ \ D/4=16\ \ ,\ \ x_1=\dfrac{1}{9}\ \ ,\ \ x_2=1\notin ODZ\\\\\\Otvet:\ x=\dfrac{1}{9}\ .$