Question

-x(в квадрате)+5x-4>/0

Answer 1

$-x^2+5x-4 \geq 0; x^2-5x+4 \leq 0;(x-1)(x-4) \leq 0;$
левая часть неравенства - парабола ветками вверх
x∈[1;4]

Answer 2

To solve the inequality -x² + 5x - 4 > 0, we can follow these steps:

Step 1: Factorize the quadratic equation if possible. In this case, we can't easily factorize it, so we'll move on to the next step.

Step 2: Find the critical points by setting the equation equal to zero. To do this, we set -x² + 5x - 4 = 0 and solve for x. However, this equation does not factor easily, so we'll use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

For -x² + 5x - 4 = 0, the coefficients are: a = -1 b = 5 c = -4

Using the quadratic formula, we have: x = (-(5) ± √((5)² - 4(-1)(-4))) / (2(-1)) x = (-5 ± √(25 - 16)) / (-2) x = (-5 ± √9) / (-2) x = (-5 ± 3) / (-2)

So we have two critical points: x1 = (-5 + 3) / (-2) = -1 x2 = (-5 - 3) / (-2) = 4

Step 3: Create intervals using the critical points. We now have three intervals: (-∞, -1), (-1, 4), and (4, +∞).

Step 4: Test a value in each interval to determine the sign of the quadratic expression. We'll choose a value from each interval and substitute it back into the inequality.

For the interval (-∞, -1), let's test x = -2: Plug x = -2 into the inequality: -(-2)² + 5(-2) - 4 > 0 4 - 10 - 4 > 0 -10 > 0 (False)

For the interval (-1, 4), let's test x = 0: Plug x = 0 into the inequality: -(0)² + 5(0) - 4 > 0 -4 > 0 (False)

For the interval (4, +∞), let's test x = 5: Plug x = 5 into the inequality: -(5)² + 5(5) - 4 > 0 -25 + 25 - 4 > 0 -4 > 0 (False)

Step 5: Analyze the signs. Since the inequality is asking for when the quadratic expression is greater than zero (positive), and none of the intervals resulted in a positive value, there are no solutions to the inequality -x² + 5x - 4 > 0.

Therefore, the inequality has no solutions.