Question

3√(x^2-5x+4)<6+5x-x^2

Answer 1

Для начала найдём ОДЗ
(x^2-5x+4)≥0
(По теореме Виета
x^2-5x+4=0
х1+х2=5 х1=1
х1*х2=4 х2=3)
Отметив промежутки знакопостоянства получается:
х(-бесконечность;1] [3; бесконечность)
Вводим переменную а
а=√(x^2-5x+4)
Подставляем
3а<-а²-10
а²+3а-10>0
(По теореме Виета
а^2+3а-10=0
х1+х2=-3 х1=-5
х1*х2=-10 х2=2)
Отметив промежутки знакопостоянства получается:
х(-бесконечность;-5)(2;+бесконечность)
Проверяем ОДЗ
Ответ:х(-бесконечность;-5) [3;+бесконечность)

Answer 2

To solve the inequality 3√(x^2 - 5x + 4) < 6 + 5x - x^2, we need to follow these steps:

Step 1: Move all the terms to one side of the inequality to obtain a quadratic expression on the right side: 3√(x^2 - 5x + 4) - 6 - 5x + x^2 < 0

Step 2: Square both sides of the inequality to eliminate the cube root: [3√(x^2 - 5x + 4) - 6 - 5x + x^2]^2 < 0

Step 3: Expand the squared expression: 9(x^2 - 5x + 4) - 36√(x^2 - 5x + 4) + 36 - 36x + 6x(x^2 - 5x + 4) + 25x^2 - 10x(x^2 - 5x + 4) + x^4 - 10x^3 + 25x^2 < 0

Step 4: Simplify the expression: 9x^2 - 45x + 36 - 36√(x^2 - 5x + 4) + 6x^3 - 30x^2 + 24x + 25x^2 - 10x^3 + x^4 - 10x^3 + 25x^2 < 0

Step 5: Combine like terms: x^4 - 14x^3 - 11x^2 - 9x + 36 - 36√(x^2 - 5x + 4) < 0

Step 6: Solve the quadratic equation x^4 - 14x^3 - 11x^2 - 9x + 36 = 0 to find the critical points. This equation might not have nice analytical solutions, so numerical methods like graphing or using a calculator can be employed.

Step 7: Plot the critical points on a number line and check the sign of the expression x^4 - 14x^3 - 11x^2 - 9x + 36 in each interval to determine the solution to the inequality.

Unfortunately, the solution process becomes more complicated due to the presence of the fourth power term and the presence of the square root. Without the specific values of x, I am unable to provide an exact solution.