Question

$\frac{2 {x}^{2} - 5x + 3 }{(10x - 5)(x - 1)} = 0$
уравнение кому не лень)

Answer 1

$x \in \left(-\infty; \frac{1}{2} \right) \cup \left( \frac{1}{2} ;1 \right) \cup \left(1;+ \infty \right)$

$\frac{2 x^2 - 5x + 3 }{(10x - 5)(x - 1)} = 0$

$2 x^2 - 5x + 3 =0$

$\Delta=(-5)^2-4 \cdot 2 \cdot 3=25-24=1$

$\sqrt{\Delta} = \sqrt{1} =1$

$x_1= \frac{5-1}{2 \cdot 2} = \frac{4}{4}=1\not \in\left(-\infty; \frac{1}{2} \right) \cup \left( \frac{1}{2} ;1 \right) \cup \left(1;+ \infty \right)$

$x_2= \frac{5+1}{2 \cdot 2} = \frac{6}{4}= \frac{3}{2}$