Question

2y+3/2y+2 - Y+1/2y-2+1/y^2-1=0

Answer 1

(2у+3/2у+2)-(у+1/2у-2)+(1/у²-1)=0 у≠-1,у≠1
(2у+3/2(у+1))-(у+1/2(у-1))+(1/(у+1)(у-1))=0
((у-2)(2у+3)-(у+1)²+2/2(у-1)(у+1))=0
(2у²+3у-2у-3-(у²+2у+1)/2(у-1)(у+1))=0
(2у²+3у-2у-3-у²-2у+-1/2(у-1)(у+1))=0
(у²-у-2/2(у-1)(у+1))=0
(у²+у-2у-2/2(у-1)(у+1))=0
(у(у+1)-2(у+1)/2(у-1)(у+1))=0
((у+1)(у-2)/2(у-1)(у+1))=0
(у-2/2(у-1))=0
у-2=0
у=2,у≠-1,у≠1
Ответ:у=2

Answer 2

To solve the equation:

(2y + 3)/(2y + 2) - (y + 1)/(2y - 2) + 1/(y^2 - 1) = 0

Let's simplify it step by step:

Step 1: Find the common denominator for all three terms. The common denominator for the three terms is (2y + 2)(2y - 2)(y^2 - 1) = 2(y + 1) * 2(y - 1) * (y + 1)(y - 1) = 4(y + 1)(y - 1)(y^2 - 1).

Step 2: Rewrite each term with the common denominator:

(2y + 3)/(2y + 2) = (2y + 3) * (2(y - 1)(y + 1)) / 4(y + 1)(y - 1)(y^2 - 1) = (2y + 3)(y - 1)(y + 1) / 2(y + 1)(y - 1)(y^2 - 1)

(y + 1)/(2y - 2) = (y + 1) * (2(y + 1)(y - 1)) / 4(y + 1)(y - 1)(y^2 - 1) = 2(y + 1)(y - 1) / 4(y + 1)(y - 1)(y^2 - 1) = 1 / 2(y^2 - 1)

1/(y^2 - 1) = 4(y + 1)(y - 1) / 4(y + 1)(y - 1)(y^2 - 1) = 1 / 4(y + 1)(y - 1)

Step 3: Substitute the rewritten terms back into the equation:

(2y + 3)(y - 1)(y + 1) / 2(y + 1)(y - 1)(y^2 - 1) - 2(y + 1)(y - 1) / 4(y + 1)(y - 1)(y^2 - 1) + 1 / 4(y + 1)(y - 1) = 0

Step 4: Eliminate common factors (y + 1)(y - 1) from the equation:

[(2y + 3)(y - 1) - 2(y + 1) + 1] / 2(y + 1)(y - 1)(y^2 - 1) = 0

Step 5: Expand and simplify:

(2y^2 + 3y - 2y - 3 - 2y - 2 + 1) / 2(y + 1)(y - 1)(y^2 - 1) = 0

(2y^2 - 6) / 2(y + 1)(y - 1)(y^2 - 1) = 0

Step 6: Divide both sides by 2 to simplify:

(y^2 - 3) / (y + 1)(y - 1)(y^2 - 1) = 0

Now, the equation is simplified. To find the values of 'y,' we can set the numerator equal to zero since a fraction is zero if and only if the numerator is zero:

y^2 - 3 = 0

Now, solve for 'y':

y^2 = 3

y = ±√3

So the solutions to the original equation are:

y = √3 and y = -√3.