Question

(6-х)²=(4y-5)²=(2a+4)²=(b+6a)²=(2y-3)(2y+3)=​

Answer 1

Ответ:

1) после =36-x в квадрате

2)8y в квадрате-25

3)4a в квадрате +16

4)b в квадрате+36a квадрате

Answer 2

Let's solve each equation step by step:

1. (6 - x)² = (4y - 5)² Expanding both sides: 36 - 12x + x² = 16y² - 40y + 25

2. (4y - 5)² = (2a + 4)² Expanding both sides: 16y² - 40y + 25 = 4a² + 16a + 16

3. (2a + 4)² = (b + 6a)² Expanding both sides: 4a² + 16a + 16 = b² + 12ab + 36a²

4. (b + 6a)² = (2y - 3)(2y + 3) Expanding both sides: b² + 12ab + 36a² = 4y² - 9

Now we have a system of equations. We can try to solve it by substitution or elimination. Let's solve it using substitution:

From equation 3, we have: 4a² + 16a + 16 = b² + 12ab + 36a² 16 + 16a = b² + 12ab + 32a² 16 + 16a = b² + 4ab + 4(8a²) 16 + 16a = (b + 2a)²

From equation 4, we have: b² + 12ab + 36a² = 4y² - 9 (b + 6a)² = 4y² - 9 (b + 6a)² + 9 = 4y²

Now, we have two equations: 16 + 16a = (b + 2a)² (equation 5) (b + 6a)² + 9 = 4y² (equation 6)

To solve this system, we need more information or additional equations. The given equations do not provide enough information to determine the specific values of x, y, a, and b.