Question

Докажите, что выражение (х-4у)(х-4у-8)+16 принимает НЕотрицательные значения при любых значениях переменной ОЧЕНЬ СРОЧНО!!!!!

Answer 1

Х^-4ху-8х-4ху+16у^+32у+16>0
x^-8xy+16y^-8x+32y+16>0
(x-4y)^-8x+32y+16>0
(x-4y)^-8(x-4y)+16>0

Answer 2

Proof that the expression (х-4у)(х-4у-8)+16 takes non-negative values for any values of the variable

To prove that the expression (х-4у)(х-4у-8)+16 takes non-negative values for any values of the variable, we can analyze the factors and constants involved in the expression.

Let's break down the expression step by step:

1. (х-4у): This is the first factor in the expression. It represents the difference between the variable х and 4 times the variable у. Since this is a quadratic term, it can take any real value.

2. (х-4у-8): This is the second factor in the expression. It represents the difference between the variable х, 4 times the variable у, and 8. Again, this is a quadratic term and can take any real value.

3. 16: This is a constant term in the expression. It is a positive value.

Now, let's consider the possible scenarios:

- If both factors, (х-4у) and (х-4у-8), are positive, then their product will be positive.

- If both factors are negative, then their product will be positive as well, since multiplying two negative numbers results in a positive number.

- If one factor is positive and the other is negative, then their product will be negative.

However, in all of these scenarios, we have the constant term 16 added to the product of the factors. Since 16 is a positive value, it will always ensure that the overall expression takes non-negative values.

Therefore, we can conclude that the expression (х-4у)(х-4у-8)+16 takes non-negative values for any values of the variable.