Question

lg(2-3x)+lg(2+3x)=lg(4-x)+lg(x) Помогите!

Answer 1

$lg(2-3x)+lg(2+3x)=lg(4-x)+lgx$
ОДЗ:
$left { {{2-3x textgreater 0} atop {2+3x textgreater 0}}}} right.$
$left { {{4-x textgreater 0} atop {x textgreater 0}} right.$

$left { {{x textless frac{2}{3} } atop {x textgreater - frac{2}{3} }}}} right.$
$left { {{x textless 4} atop {x textgreater 0}} right.$
$x$ $(0; frac{2}{3} )$

$lg[(2-3x)(2+3x)]=lg[(4-x)x]$
$lg(4-9x^2)=lg(4x-x^2)$
$4-9x^2=4x-x^2$
$8x^2+4x-4=0$
$2x^2+x-1=0$
$D=1^2-4*2*(-1)=9$
$x_1= frac{-1+3}{4}=0.5$
$x_2= frac{-1-3}{4}=-1$  ∅

Ответ: 0.5