Question

X^2-2020x+2020^2020<0

Answer 1

Ответ:

нет решений

Пошаговое объяснение:

$x^2-2020x+2020^{2020}$

Парабола выше оси x, значит решений у неравенства нет.

Answer 2

To solve the inequality $x^2 - 2020x + 2020^{2020} < 0$, we can find the critical points and analyze the intervals.

The given quadratic inequality can be factored as:

$(x - r_1)(x - r_2) < 0$

Where $r_1$ and $r_2$ are the roots of the quadratic equation $x^2 - 2020x + 2020^{2020} = 0$.

To find the roots, we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For our equation, $a = 1$, $b = -2020$, and $c = 2020^{2020}$.

Substituting these values into the quadratic formula:

$x = \frac{2020 \pm \sqrt{2020^2 - 4 \cdot 1 \cdot 2020^{2020}}}{2 \cdot 1}$

Simplifying:

$x = \frac{2020 \pm \sqrt{2020^2 - 4 \cdot 2020^{2022}}}{2}$

$x = \frac{2020 \pm 2020\sqrt{1 - 4 \cdot 2020^{2018}}}{2}$

$x = 1010 \pm 1010\sqrt{1 - 4 \cdot 2020^{2018}}$

Since the discriminant $b^2 - 4ac$ is nonnegative, the roots are real.

The roots are:

$r_1 = 1010 - 1010\sqrt{1 - 4 \cdot 2020^{2018}}$

$r_2 = 1010 + 1010\sqrt{1 - 4 \cdot 2020^{2018}}$

To determine the intervals where the inequality is true, we need to consider the signs of the quadratic expression $(x - r_1)(x - r_2)$ in different intervals.

Let's consider the following cases:

1. When $x < r_1$: $(x - r_1)$ and $(x - r_2)$ are both negative, so their product is positive. Therefore, the inequality is not satisfied in this interval.

2. When $r_1 < x < r_2$: $(x - r_1)$ is positive, and $(x - r_2)$ is negative, so their product is negative. Therefore, the inequality is satisfied in this interval.

3. When $x > r_2$: $(x - r_1)$ and $(x - r_2)$ are both positive, so their product is positive. Therefore, the inequality is not satisfied in this interval.

Hence, the solution to the inequality $x^2 - 2020x + 2020^{2020} < 0$ is:

$r_1 < x < r_2$