Допоможіть будь-ласка

Question 1

Question 2

$( \frac{5x}{x-10}+ \frac{20x}{x^2-20x+10} ): \frac{4x-24}{x^2-100} - \frac{25x}{x-10}= \frac{5x}{4}$

Решение:
$5x( \frac{1}{x-10}+ \frac{4}{x^2-20x+10} )* \frac{x^2-100}{4(x-6)} - \frac{25x}{x-10}= \frac{5x}{4}$

$5x* \frac{x^2-20x+10+4(x-10)}{(x-10)(x^2-20x+10)}* \frac{(x-10)(x+10)}{4(x-6)} - \frac{25x}{x-10}= \frac{5x}{4}$

$5x* \frac{x^2-16x-30}{x^2-20x+10}* \frac{x+10}{4(x-6)} - \frac{25x}{x-10}= \frac{5x}{4}$

Умножим обе части уравнения на 4/5
$x* \frac{x^2-16x-30}{x^2-20x+10}* \frac{x+10}{(x-6)} - \frac{20x}{x-10}= x$

$x* \frac{x^2-16x-30}{x^2-20x+10}* \frac{x+10}{(x-6)} - \frac{20x}{x-10}- x =0$

$x(\frac{x^2-16x-30}{x^2-20x+10}* \frac{x+10}{(x-6)} - 1 -\frac{20}{x-10}) =0$

$x(\frac{(x^2-16x-30)(x+10)-(x^2-20x+10)(x-6)}{(x^2-20x+10)(x-6)} -\frac{20}{x-10}) =0$

$x(\frac{x^3-6x^2-190x-300-(x^3-26x^2+130x-60)}{(x^2-20x+10)(x-6)} -\frac{20}{x-10}) =0$

$x(\frac{20x^2-320x-240}{(x^2-20x+10)(x-6)} -\frac{20}{x-10}) =0$

$x(\frac{x^2-16x-12}{(x^2-20x+10)(x-6)} -\frac{1}{x-10}) =0$

$x*\frac{(x^2-16x-12)(x-10)-(x^2-20x+10)(x-6)}{(x^2-20x+10)(x-6)(x-10)} =0$

$x*\frac{x^3-26x^2+148x+120-(x^3-26x^2-130x-60)}{(x^2-20x+10)(x-6)(x-10)} =0$

$x*\frac{18x+180}{(x^2-20x+10)(x-6)(x-10)} =0$

$x*\frac{18(x+10)}{(x^2-20x+10)(x-6)(x-10)} =0$

$\frac{x(x+10)}{(x^2-20x+10)(x-6)(x-10)} =0$

x=0, x=-10