Question

1) решить неравенство: а) х^2-0,04<=0 б) 0,01х^2-9>0 в) х^2-3х-4<0 г) 3х^2-4х+8>=0 д) -х^2+3х-5>0 е) х^2+20х+100<=02) Решить методом интервалов неравенство:х (х-1)(х+2)>=0 ПОМОГИТЕ ПОЖАЛУЙСТА!

Answer 1

2)
x(x-1)(x+2)>=0
x1=0     x2=1      x3=-2

x принадлежит [-2;0]  [1;+бесконечности)

Answer 2

1) Решение неравенств:

а) х^2 - 0,04 <= 0

To solve this inequality, we can find the values of x that make the expression on the left side less than or equal to zero.

The given inequality can be rewritten as:

x^2 <= 0.04

Taking the square root of both sides, we get:

x <= ±0.2

Therefore, the solution to the inequality is x <= -0.2 or x >= 0.2. [[1]]

б) 0.01x^2 - 9 > 0

To solve this inequality, we can find the values of x that make the expression on the left side greater than zero.

The given inequality can be rewritten as:

0.01x^2 > 9

Dividing both sides by 0.01, we get:

x^2 > 900

Taking the square root of both sides, we get:

x > ±30

Therefore, the solution to the inequality is x > -30 or x < 30. [[2]]

в) x^2 - 3x - 4 < 0

To solve this inequality, we can find the values of x that make the expression on the left side less than zero.

The given inequality can be factored as:

(x - 4)(x + 1) < 0

To determine the sign of the expression, we can use a sign chart or test values from each interval.

Testing a value less than -1, such as x = -2, we get:

(-2 - 4)(-2 + 1) < 0 (-6)(-1) < 0 6 < 0 (False)

Testing a value between -1 and 4, such as x = 0, we get:

(0 - 4)(0 + 1) < 0 (-4)(1) < 0 -4 < 0 (True)

Testing a value greater than 4, such as x = 5, we get:

(5 - 4)(5 + 1) < 0 (1)(6) < 0 6 < 0 (False)

Therefore, the solution to the inequality is -1 < x < 4. [[3]]

г) 3x^2 - 4x + 8 >= 0

To solve this inequality, we can find the values of x that make the expression on the left side greater than or equal to zero.

The given inequality can be factored as:

(3x - 2)(x - 4) >= 0

To determine the sign of the expression, we can use a sign chart or test values from each interval.

Testing a value less than 2/3, such as x = 0, we get:

(3(0) - 2)(0 - 4) >= 0 (-2)(-4) >= 0 8 >= 0 (True)

Testing a value between 2/3 and 4, such as x = 1, we get:

(3(1) - 2)(1 - 4) >= 0 (1)(-3) >= 0 -3 >= 0 (False)

Testing a value greater than 4, such as x = 5, we get:

(3(5) - 2)(5 - 4) >= 0 (13)(1) >= 0 13 >= 0 (True)

Therefore, the solution to the inequality is x <= 2/3 or x >= 4. [[4]]

д) -x^2 + 3x - 5 > 0

To solve this inequality, we can find the values of x that make the expression on the left side greater than zero.

The given inequality can be rewritten as:

x^2 - 3x + 5 < 0

Since the coefficient of x^2 is negative, the parabola opens downwards. Therefore, the inequality is satisfied between the roots of the quadratic equation.

Using the quadratic formula, we can find the roots:

x = (3 ± √(3^2 - 4(1)(5))) / (2(1)) x = (3 ± √(9 - 20)) / 2 x = (3 ± √(-11)) / 2

Since the discriminant is negative, there are no real roots. Therefore, the inequality is not satisfied for any real values of x.

Therefore, there is no solution to the inequality. [[5]]

е) x^2 + 20x + 100 <= 0

To solve this inequality, we can find the values of x that make the expression on the left side less than or equal to zero.

The given inequality can be factored as:

(x + 10)^2 <= 0

Since the square of a real number is always greater than or equal to zero, the only way for the inequality to hold is if the expression is equal to zero.

Therefore, the solution to the inequality is x = -10. [[6]]

2) Решение методом интервалов:

х + (х - 1)(х + 2) >= 0

To solve this inequality using the interval method, we can find the intervals where the expression on the left side is greater than or equal to zero.

First, let's find the critical points by setting each factor equal to zero:

x = 0 (from x) x - 1 = 0 (from (x - 1)) x + 2 = 0 (from (x + 2))

The critical points are x = 0, x = 1, and x = -2.

Next, we can create an interval chart to determine the sign of the expression in each interval:

Interval (-∞, -2): (-)(-)(-) = - Interval (-2, 0): (-)(-)(+) = + Interval (0, 1): (+)(-)(+) = - Interval (1, +∞): (+)(+)(+) = +

Therefore, the solution to the inequality is x <= -2 or 0 <= x <= 1. [[7]]

I hope this helps! Let me know if you have any further questions.