Question

Помогите пожалуйста решить неравенства. 1. a) 4-7x>15 b) 3y-8<9+3y c) 12-5m<= 1-2(6+4m) d) 7+5x/6>= 3+15-4x/8 e) x^2-15x+36>0 2. 2x+1/3-x<=0 3. Решить,используя метод интервалов. 2x+8/x^2-49>=0

Answer 1

1)
а) х <-11
б) все числа
с)(-бескон; 3) и (12;+бескон)
3)(-7;-4) и (7;+бескон.)

Answer 2

a) 4-7x>15 -7x>11 x<-11/7

b) 3y-8<9+3y -8<9 This inequality is true for all values of y.

c) 12-5m<=1-2(6+4m) 12-5m<=1-12-8m 3m<=-23 m<=-23/3

d) 7+5x/6>=3+15-4x/8 7+5x/6>=18-4x/8 7+5x/6>=18-x/2 5x/6+x/2>=11 x/6>=11-5x/2 x>=66-15x 16x>=66 x>=33/8

e) x^2-15x+36>0 (x-3)(x-12)>0 This inequality is true when x<3 or x>12, so the solution is x<3 or x>12.

2. 2x+1/3-x<=0 5x-1<=0 x<=1/5

3. To solve 2x+8/x^2-49>=0 using interval method, we can first find the critical points of the function. The critical points are the values of x where the function is either equal to zero or undefined.

Setting the numerator equal to zero, we get: 2x + 8 = 0 x = -4

Setting the denominator equal to zero, we get: x^2 - 49 = 0 (x + 7)(x - 7) = 0 x = -7 or x = 7

Thus, the critical points are x = -7, x = -4, and x = 7. We can use these critical points to divide the x-axis into four intervals: (-∞, -7), (-7, -4), (-4, 7), and (7, ∞).

Next, we need to determine the sign of the expression 2x+8/x^2-49 in each interval. We can use test points within each interval to do this. For example, in the interval (-∞, -7), we can use the test point x = -8: 2(-8) + 8 / (-8)^2 - 49 = -8/33 < 0

Similarly, we can find the signs of the expression in the other intervals: (-7, -4): 2(-5) + 8 / (-5)^2 - 49 = 6/121 > 0 (-4, 7): 2(0) + 8 / (0)^2 - 49 is undefined (7, ∞): 2(8) + 8 / (8)^2 - 49 = 8/33 > 0

Therefore, the solution set is the union of the intervals where the expression is nonnegative: (-∞, -7) U (7, ∞)