Question

(2y-3)(3y+1)+2(y-5)(y+5)=2(1-2y)²+6y

Answer 1

Ответ:

(2y-3)(3y+1)+2(y-5)(y+5)=2(1-2y)²+6y,

6y2+2y-9y-3+2y2+10y-10y-50= =2-4y-4y+8y2+6y,

-5y-55 = 0,

5y+55 = 0,

5y = -55,

y = -11.

Answer 2

(2y-3)(3y+1)+2(y-5)(y+5)=2(1-2y)²+6y
6у²+2у-9у-3+2(у²-25)=2(1-4у+4у²)+6у
6у²-7у-3+2у²-50=2-8у+8у²+6у
8у²-7у-53=8у²-2у+2
8у²-7у-8у²+2у=2+53
-5у=55|÷(-5)
у=-11.

Проверка:

(2×(-11)-3)(3×(-11)+1)+2((-11)-5)((-11)+5)=2(1-2×(-11))²+6(-11)
(-22-3)(-33+1)+2((-16))(-6)=2(1+22)²+(-66)
(-25)×(-32)+2×96=2×23²-66
800+192=2×529-66
992=1058-66
992=992-истина

Answer 3

Let's simplify the given equation step by step:

(2y - 3)(3y + 1) + 2(y - 5)(y + 5) = 2(1 - 2y)² + 6y

First, let's expand the expressions on both sides of the equation:

(2y - 3)(3y + 1) = 2(1 - 2y)² + 6y

Now, expand each expression:

(2y)(3y) + (2y)(1) + (-3)(3y) + (-3)(1) + 2(y)(y) + 2(y)(5) + 2(5)(y) + 2(5)(5) = 2(1 - 2y)² + 6y

Now, simplify each term:

6y² + 2y - 3y - 3 + 2y² + 10y + 10y + 50 = 2(1 - 2y)² + 6y

Combine like terms on both sides of the equation:

8y² + 16y + 47 = 2(1 - 2y)² + 6y

Now, let's simplify the right side by squaring (1 - 2y):

(1 - 2y)² = 1 - 4y + 4y²

So, the equation becomes:

8y² + 16y + 47 = 2(1 - 4y + 4y²) + 6y

Now, distribute the 2 on the right side:

8y² + 16y + 47 = 2 - 8y + 8y² + 6y

Now, combine like terms on the right side:

8y² + 16y + 47 = 2 + 6y² - 2y

Now, let's move all terms to one side to set the equation equal to zero:

8y² + 16y + 47 - 2 - 6y² + 2y = 0

Combine like terms:

(8y² - 6y²) + (16y + 2y) + (47 - 2) = 0

Now, simplify:

2y² + 18y + 45 = 0

Now, we have a quadratic equation in the form of ay² + by + c = 0. To solve for y, we can use the quadratic formula:

y = (-b ± √(b² - 4ac)) / (2a)

In this case, a = 2, b = 18, and c = 45. Plug these values into the quadratic formula:

y = (-18 ± √(18² - 4(2)(45))) / (2(2))

Now, calculate the discriminant (the value inside the square root):

Discriminant = 18² - 4(2)(45) = 324 - 360 = -36

Since the discriminant is negative, the quadratic equation has no real solutions. It means that there are no real values of y that satisfy the original equation.