Question

У выражение (3y^2-10y+8)/(4y^2-36)∙(y-3)/(y-2)+(0,25-y)/(y+3)

Answer 1

ответ: -1/4

Объяснение:

$\frac{3у^2-4y-6y=8}{4(y^2-9)} *\frac{y-3}{y-2} +\frac{\frac{1}{4}-y }{y+3}$

$\frac{y*(3y-4)-2(3y-4)}{4(y-3)*(y+3)} *\frac{y-3}{y-2} +\frac{\frac{1-4y}{4} }{y+3}$

$\frac{(3y-4)*(y-2)}{4(y+3)} *\frac{1}{y-2} +\frac{1-4y}{4(y+3)\\}$

$\frac{3y-4}{4(y+3)} +\frac{1-4y}{4(y+3)}$

$\frac{3y-4+1-4y}{4(y+3)}$

$\frac{-y-3}{4(y+3)}$

$\frac{-(y+3)}{4(y+3)}$

-1/4