Question

(0.1) ^4x²+3x ≤ (0,1)^3x²+4

Answer 1

Ответ: (-∞; -4] ∪ [1; +∞).

Пошаговое объяснение:

Т.к. показательная функция у = (0,1)ˣ убывает, то перейдем к неравенству

4х² + 3х ≥ 3х² + 4,

х² + 3х - 4 ≥ 0.

Найдем нули функции у = х² + 3х - 4:

х² + 3х - 4 = 0,

D = 3² - 4 · 1 · (-4) = 9 + 16 = 25; √25 = 5

х₁ = (-3 + 5)/(2 · 1) = 2/2 = 1,

х₂ = (-3 - 5)/(2 · 1) = -8/2 = -4.

Решим неравенство методом интервалов:

    +            -              +

-----------|-----------|--------------->

           -4           1

х ∈ (-∞; -4] ∪ [1; +∞).

Answer 2

To solve the inequality $(0.1)^{4x^2 + 3x} \leq (0.1)^{3x^2 + 4}$, we can simplify the expression using the properties of exponents and then solve for $x$.

Let's start by simplifying the inequality using the properties of exponents:

$(0.1)^{4x^2 + 3x} \leq (0.1)^{3x^2 + 4}$

Now, we can use the property of exponents that says when you have the same base raised to different exponents and they are being compared, the exponents themselves can be compared:

$4x^2 + 3x \leq 3x^2 + 4$

Next, let's rearrange this inequality to isolate $x$:

$4x^2 + 3x - 3x^2 - 4 \leq 0$

Combine like terms:

$x^2 + 3x - 4 \leq 0$

Now, let's factor the quadratic expression:

$(x + 4)(x - 1) \leq 0$

We now have the inequality factored. To solve for $x$, we need to find the intervals where this expression is less than or equal to zero. This occurs when $x$ is between -4 and 1:

$-4 \leq x \leq 1$

So, the solution to the inequality $(0.1)^{4x^2 + 3x} \leq (0.1)^{3x^2 + 4}$ is $-4 \leq x \leq 1$.