Question

Решить неравенство(2x – 1)(2х^2 – 3х + 5) < 0​

Answer 1

$\displaystyle \tt (2x-1)(2x^2-3x+5)$

Answer 2

(2х-1)(2х²-3х+5)<0

отрицательное число получается тогда как перемножаются положительное и отрицательное число

2х-1>0                                    D=3²-4×2×5=-31 ⇒ корней нет

2x>1                                       а учитывая а=2>0 ⇒ 2х²-3х+5>0

x ∈(0.5;∞)

значит рассматриваем второй вариант

2х²-3х+5>0                                                    2x-1<0

как мы уже выяснили                                 2x<1

x∈(-∞;∞)                                                            x∈(-∞;0.5)

(-∞;∞) ∩ (-∞;0.5) = (-∞;0.5)

ответ:  х ∈ (-∞;0.5)

Answer 3

To solve the inequality $(2x - 1)(2x^2 - 3x + 5) < 0$, we need to find the intervals of $x$ for which the expression is negative.

To do this, we'll follow these steps:

Step 1: Find the critical points by setting each factor to zero and solving for $x$. Step 2: Create a sign chart to determine the sign of each factor in the inequality. Step 3: Find the intervals where the expression is negative.

Let's proceed with the calculations:

Step 1: Find the critical points: $(2x - 1) = 0$ ⇒ $x = \frac{1}{2}$

$(2x^2 - 3x + 5) = 0$ ⇒ This quadratic doesn't have real roots because its discriminant $b^2 - 4ac = (-3)^2 - 4(2)(5) = 9 - 40 = -31$ is negative. So, there are no additional critical points.

Step 2: Create a sign chart: We'll evaluate the sign of each factor for three intervals: $x < \frac{1}{2}$, $x = \frac{1}{2}$, and $x > \frac{1}{2}$.

markdownCopy code
Interval           |  (2x - 1)  | (2x^2 - 3x + 5)
-------------------------------------------------
x < 1/2            |     -     |        +
x = 1/2            |     0     |        +
x > 1/2            |     +     |        +

Step 3: Find the intervals where the expression is negative: The inequality is satisfied when the product of the two factors is negative, which means we need one factor to be negative and the other factor to be positive.

The solution is given by: $\text{Solution: } \frac{1}{2} > x$

Thus, the inequality $(2x - 1)(2x^2 - 3x + 5) < 0$ is satisfied for $x$ in the interval $-\infty < x < \frac{1}{2}$.