Question

5у^2-20у+20/3у^2+3у+3 × у^3-1/10у^2 -40

Answer 1

Ответ:

$...= \frac{(y - 2)(y - 1)}{6(y + 2)} \\ _ {или}\\ ... \: = \: \frac{y^2- 3y + 2}{6y + 12}$

Объяснение:

$\frac{5y^2-20y+20}{3y^2+3y+3} \cdot \frac{ y^3-1}{10y^2 -40} \: = \\ = \frac{5(y^2-4y+4)}{3(y^2+y+1)} \cdot \frac{ y^3-1}{10(y^2 -4)} \: = \\ = \frac{5(y - 2)^2}{3(y^2+y+1)} \cdot \frac{ (y-1)(y^2 + y + 1)}{10(y -2)(y + 2)} \: = \\ = \frac{ \cancel{5} (\cancel{y - 2})(y - 2)}{3( \cancel{y^2+y+1})} \cdot \frac{ (y-1) (\cancel{y^2 + y + 1})}{2\cdot \cancel{5} (\cancel{y -2})(y + 2)} \: = \\ = \frac{(y - 2)(y - 1)}{6(y + 2)} = \frac{y^2- 3y + 2}{6y + 12}$