Question

Нужно решить уравнение
2x^2-3x-20/6x^2-20x-16=(6x-4)^2/36x^2-16

Answer 1

$frac{2x^2-3x-20}{6x^2-20x-16} =frac{(6x-4)^2}{36x^2-16} \frac{2x^2-3x-20}{2(3x^2-10x-8)} =frac{(6x-4)^2}{(6x-4)(6x+4)}$

2x² - 3x - 20 = 0
D = 9 + 160 = 169
$x_1= frac{3-13}{4}= -frac{5}{2} \ x_2= frac{3+13}{4}=4$
2x² - 3x - 20 = 2·(x + 5/2)·(x - 4)

3x² - 10x - 8 = 0
D = 100 + 96 = 196
$x_1= frac{10-14}{6}= -frac{2}{3} \ x_2= frac{10+14}{6}=4$
3x² - 10x - 8 = 3·(x + 2/3)·(x - 4)

$frac{2(x+ frac{5}{2})(x-4)}{2*3(x+ frac{2}{3})(x-4)} =frac{(6x-4)^2}{(6x-4)(6x+4)}\ frac{(2x+5)(x-4)}{(6x+4)(x-4)} =frac{(6x-4)^2}{(6x-4)(6x+4)} \ \ frac{2x+5}{6x+4} =frac{6x-4}{6x+4} \ x-4 neq 0 \ 6x-4 neq 0 \ \2x+5=6x-4\ x-4 neq 0 \ 6x-4 neq 0 \ 6x+4 neq 0 \ \4x=9\ x neq 4 \ 6x neq 4 \ 6x neq -4 \ \x= frac{9}{4} \ x neq 4 \ x neq frac{2}{3} \ x neq -frac{2}{3} \ \ x= frac{9}{4}$

Answer 2

Большое спасибо)