Question

X^2(3x-2)(x-8)<0 кто может решите плизз срочняк надо

Answer 1

Решим методом интервалов
х²(3х-2)(х-8)=0
уравнение равно нулю когда хотя бы один из множителей равен нулю
х²=0             или      3х-2=0                или             х-8=0
х=0               или       х=2/3                 или             х=8
    -            +      -                 +
           .        .          .                                  точки должны быть не закрашены т.к. строго меньше
           0       2/3      8 
Ответ: х∈(-бесконечности;0) и (2/3;8)

Answer 2

I can help you solve this quadratic inequality. First, you need to factor the expression X^2(3x-2)(x-8) and find the values of x that make it equal to zero. These are x = 0, x = 2/3, and x = 8. Then, you need to plot these values on a number line and divide it into four intervals: (-∞, 0), (0, 2/3), (2/3, 8), and (8, ∞). Next, you need to test each interval by plugging in any value from that interval into the expression and checking if it is less than zero. For example, if you plug in x = -1 into X^2(3x-2)(x-8), you get (-1)^2(3(-1)-2)(-1-8) = 99, which is greater than zero. This means that the interval (-∞, 0) is not part of the solution. You can repeat this process for the other intervals and find that the only intervals that satisfy the inequality are (0, 2/3) and (8, ∞). Therefore, the solution is x ∈ (0, 2/3) ∪ (8, ∞). You can also see a graphical representation of the solution here. I hope this helps you understand how to solve quadratic inequalities. If you have any other questions, feel free to ask me.

: https://www.symbolab.com/solver/inequality-calculator/X%5E%7B2%7D%5Cleft(3x-2%5Cright)%5Cleft(x-8%5Cright)%3C0