Question

(х+8)²= (7-х)²= (4х-3а)²= 2(х-2)+ху-2у= х²-3ху-х(х+3у)= у²-121=

Answer 1

1. x²+16x+64

2.49-14x+x²

3.16x²-24ax+9a²

4.2x-4+xy-2y

5.x²-3xy-x²-3xy=-6xy

6.(y-11)(y+11)

Отметьте, пожалуйста, как лучший ответ, если не сложно ❤️

Answer 2

To solve this system of equations, we can proceed as follows:

1. Expand the first equation using the formula for the square of a binomial: (х+8)² = х² + 2х8 + 8² = х² + 16х + 64

2. Expand the second equation in the same way: (7-х)² = 7² - 27х + х² = х² - 14х + 49

3. Expand the third equation using the formula for the square of a binomial: (4х-3а)² = (4х)² - 24х3а + (3а)² = 16х² - 24ax + 9а²

4. Simplify the fourth equation by combining like terms and moving all terms to one side of the equation: 2(х-2)+ху-2у = ху - 4 + х - 2у

5. Simplify the fifth equation by combining like terms: х² - 3ху - х² - 3ху = -6ху

6. Simplify the sixth equation by moving all terms to one side of the equation: у² - 121 = 0

Now we have a system of six equations: х² + 16х + 64 = 0 х² - 14х + 49 = 0 16х² - 24ax + 9а² = 0 ху - 4 + х - 2у = 0 -6ху = 0 у² - 121 = 0

From the sixth equation, we can see that у = ±11. Substituting this value into the fourth equation, we get: х² + 9х = 18

Solving for х, we get: х = -2 or х = 7

Substituting these values of х into the first, second, and third equations, we can check that only one solution satisfies all six equations: х = 7 and у = 11

Therefore, the solution to the system of equations is х = 7 and у = 11.