Question

2/y+1 + 5/y-1 - 5/y+2

Answer 1

(2(у-1)(у+2)+ 5*(у+1)(у+2) - 5(у²-1))/((у²-1)(у+2))=

(2(у²+у-2)+ 5*(у²+3у+2) - 5у²+5)/((у²-1)(у+2))=

(2у²+2у-4+ 5у²+15у+10 - 5у²+5)/((у²-1)(у+2))=(2у²+17у+11)/((у²-1)(у+2))

Answer 2

Объяснение:

Вот думаю что правильно........

Answer 3

The expression you've provided is a combination of three fractions:

1. $\frac{2}{y+1}$
2. $\frac{5}{y-1}$
3. $\frac{5}{y+2}$

To simplify the expression, you can first find a common denominator for all three fractions and then combine them.

The common denominator for these three fractions would be $(y+1)(y-1)(y+2)$.

Now, let's rewrite each fraction with this common denominator:

1. $\frac{2}{y+1} = \frac{2(y-1)(y+2)}{(y+1)(y-1)(y+2)}$
2. $\frac{5}{y-1} = \frac{5(y+1)(y+2)}{(y+1)(y-1)(y+2)}$
3. $\frac{5}{y+2} = \frac{5(y+1)(y-1)}{(y+1)(y-1)(y+2)}$

Now, add these fractions:

$\frac{2(y-1)(y+2)}{(y+1)(y-1)(y+2)} + \frac{5(y+1)(y+2)}{(y+1)(y-1)(y+2)} - \frac{5(y+1)(y-1)}{(y+1)(y-1)(y+2)}$

Combine the numerators:

$\frac{2(y-1)(y+2) + 5(y+1)(y+2) - 5(y+1)(y-1)}{(y+1)(y-1)(y+2)}$

Expand the terms:

$\frac{2y^2 + 2 - 5y^2 - 5 + 5y^2 - 1}{(y+1)(y-1)(y+2)}$

Combine like terms:

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

So, the simplified expression is $\frac{-4}{(y+1)(y-1)(y+2)}$.