Question

27a(4a-4)= (2y+c)(3y-c)= помогите решить задачу

Answer 1

1) 27a ( 4a - 4) = 108a^2 - 108a
2) (2y+c)(3y - c) = 6y^2 - 2yc+c3y+c^2=6y^2+yc+c^2=6y^2+yc - c^2 =
6y^2 - c^2+cy

Answer 2

Solving the Equation

To solve the equation 27a(4a-4) = (2y+c)(3y-c), we can follow these steps:

1. Expand and simplify the left-hand side of the equation. 2. Expand and simplify the right-hand side of the equation. 3. Set the two sides equal to each other and solve for the variables.

Step 1: Expand and Simplify the Left-hand Side

The left-hand side of the equation is 27a(4a-4). We can expand and simplify this expression.

Expanding: 27a(4a-4) = 108a^2 - 108a

Step 2: Expand and Simplify the Right-hand Side

The right-hand side of the equation is (2y+c)(3y-c). We can expand and simplify this expression.

Expanding: (2y+c)(3y-c) = 6y^2 - 2cy + 3cy - c^2 = 6y^2 + y(c - c^2)

Step 3: Set the Two Sides Equal and Solve for Variables

Now, we set the left-hand side equal to the right-hand side and solve for the variables.

Equating the expressions: 108a^2 - 108a = 6y^2 + y(c - c^2)

This equation can be solved further based on the specific values of 'a', 'y', 'c', and 'c^2'.

If you have specific values for 'a', 'y', 'c', and 'c^2', please provide them so that I can solve the equation accordingly.