Вопрос в картинках...

$log_6(2x^2-7x+6)-log_6(x-2)=log_6x$
ОДЗ:
$\left \{ {{2x^2-7x+6\ \textgreater \ 0} \atop {x-2\ \textgreater \ 0}}\atop {x\ \textgreater \ 0}} \right.$
$\left \{ {{2(x-2)(x-1.5)\ \textgreater \ 0} \atop {x\ \textgreater \ 2}}\atop {x\ \textgreater \ 0}} \right.$
$2x^2-7x+6=0$
$D=(-7)^2-4*2*6=1$
$x_1= \frac{7+1}{4}=2$
$x_2= \frac{7-1}{4}=1.5$

-------+------(1.5)------ - --------(2)-----+---------
//////////////// ////////////////

///////////////
----------(0)------------------------(2)---------------
//////////////////////////////////////////////
$x$ ∈ $(2; +$ ∞ $)$

$log_6(2x^2-7x+6)=log_6x+log_6(x-2)$
$log_6(2x^2-7x+6)=log_6[x(x-2)]$
$log_6(2x^2-7x+6)=log_6(x^2-2x)$
$2x^2-7x+6=x^2-2x$
$2x^2-7x+6-x^2+2x=0$
$x^2-5x+6=0$
$D=(-5)^2-4*1*6=1$
$x_1= \frac{5+1}{2} =3$
$x_2= \frac{5-1}{2} =2$ ∅

Ответ: 3