Question

Алгебра 7-8
Нужны все кроме 1-Го. Можно и по очереди. 2 первых упростить, 3 уравнение.
6x^2-3xy+4x-2y/9x^2+12x+4
x(y-z)-y(x-z)/x(y-z)^2-y(x-z)^2
Уравнение
4(y+9)/5y^5-45 + y+3/5y^2-15y = y^2+3y

Answer 1

$\ frac{6x^2-3xy+4x-2y}{9x^2+12x+4} \ 9x^2+12x+4=0 \ D=12^2-4*9*4=144-144=0 \ x= frac{-12}{9*2}= -frac{2}{3} \ (3x+2 )\ frac{(6x^2+4x)-(3xy+2y)}{3x+2}= frac{2x(3x+2)-y(3x+2)}{3x+2}= \ =frac{(2x-y)(3x+2)}{3x+2}=2x-y$

$frac{x(y-z)-y(x-z)}{x(y-z)^2-y(x-z)^2}= frac{xy-xz-xy+yz}{x(y^2-2yz+z^2)-y(x^2-2xz+z^2)}= \ = frac{-xz+yz}{xy^2-2xyz+xz^2-x^2y+2xyz-yz^2}= frac{-z(x-y)}{xy^2+xz^2-x^2y-yz^2}= \ = frac{-z(x-y)}{(xz^2-yz^2)-(x^2y-xy^2)}= frac{-z(x-y)}{z^2(x-y)-xy(x-y)}= \ = frac{-z(x-y)}{(z^2-xy)(x-y)}=- frac{z}{z^2-xy}$