Question

Lg (9-x)+2 lg 1/x+3=0

Answer 1

$\lg (9 - x) + 2\lg \dfrac{1}{x}+3 = 0$

ОДЗ: $\left\{\begin{array}{ccc}9 - x> 0\\\dfrac{1}{x} > 0 \ \ \ \ \ \\\end{array}\right \ \ \ \ \ \ \left\{\begin{array}{ccc}x < 9\\x > 0 \\\end{array}\right \Rightarrow x \in (0; \ 9)$

$\lg (9 - x) + \lg \dfrac{1}{x^{2}} + \lg 1000 = 0$

$\lg \dfrac{1000(9-x)}{x^{2}} = \lg 1$

$\dfrac{9000 - 1000x}{x^{2}} = 1$

$9000 - 1000x = x^{2}$

$x^{2} + 1000x - 9000 = 0$

$D = 1000^{2} - 4 \cdot 1 \cdot (-9000)= 1000000 + 36000 = 1036000$

$x_{1} = \dfrac{-1000 + 20\sqrt{2590}}{2} = -500 + 10\sqrt{2590}$

$x_{2} = \dfrac{-1000 - 20\sqrt{2590}}{2} = -500 - 10\sqrt{2590} \notin (0; \ 9)$

Ответ: $x = -500 + 10\sqrt{2590}$