Question

(1/3)^x^2 - 4x + 3 (1/3)^x-1​

Answer 1

$(\frac{1}{3})^{x^{2}-4x+3}\geq(\frac{1}{3})^{x-1}\\\\0$

       +                   -                      +

________[1]_________[4]________

                 //////////////////////

Ответ : x ∈ [1 ; 4]

Answer 2

The expression you provided is:

(1/3)^(x^2) - 4x + 3 * (1/3)^(x-1)

To simplify it further, we can break it down into separate terms:

Term 1: (1/3)^(x^2) Term 2: -4x Term 3: 3 * (1/3)^(x-1)

Let's analyze each term individually.

Term 1: (1/3)^(x^2)

The base of the term is 1/3 raised to the power of x^2. We can rewrite this term using exponentiation rules as follows:

(1/3)^(x^2) = (3^-1)^(x^2) = 3^(-x^2)

Term 2: -4x

This is a linear term with a coefficient of -4.

Term 3: 3 * (1/3)^(x-1)

The base of this term is 1/3 raised to the power of (x-1). Similar to Term 1, we can rewrite this term as:

3 * (1/3)^(x-1) = 3 * (3^-1)^(x-1) = 3 * 3^(-(x-1)) = 3 * 3^(-x+1)

Now, we can rewrite the original expression with the simplified terms:

(1/3)^(x^2) - 4x + 3 * (1/3)^(x-1) = 3^(-x^2) - 4x + 3 * 3^(-x+1)

Please note that while the expression has been simplified, it may still be further manipulated or evaluated depending on the context or specific requirements.