Question

Найти x+2y (x-4)^2+(x-y^2)^2=0

Answer 1

Ответ:

Т.к. (x-4)^2 >=0 и (x-y^2)^2 >=0 (т.е. оба слогаемые неотрицательны), то их сумма может равняться 0 только в том случае, когда оба слогаемые равны 0. Из этого следует, что х=4. Тогда у=2 либо у=-2. Значит выражение х+2у может равняться 8 либо 0.

Объяснение:

Answer 2

To find the values of x and y that satisfy the equation (x-4)^2 + (x-y^2)^2 = 0, we need to solve the system of equations obtained by expanding and equating the expression to zero.

Step 1: Expand the equation: (x-4)^2 + (x-y^2)^2 = 0 (x-4)(x-4) + (x-y^2)(x-y^2) = 0 x^2 - 8x + 16 + x^2 - 2xy^2 + y^4 = 0

Step 2: Combine like terms and arrange the equation in standard form: 2x^2 - 8x + y^4 - 2xy^2 + 16 = 0

Step 3: Set each term to zero, since the entire equation equals zero: 2x^2 - 8x + y^4 - 2xy^2 + 16 = 0 2x^2 - 8x + y^4 - 2xy^2 = -16

Step 4: Now, we have two equations:

1. 2x^2 - 8x + y^4 - 2xy^2 = -16
2. x + 2y = ?

Unfortunately, we do not have enough information to determine the exact value of x + 2y. The system of equations is not uniquely solvable because there is only one equation (equation 1) with two variables (x and y). To find specific values for x and y, we would need another equation involving both x and y. As it stands, there are infinitely many potential solutions for x and y that satisfy equation 1, but the value of x + 2y cannot be uniquely determined.