Question

1) x²<1 2) 6x-7)(5x+1)/x-3=>0 Заранее спасибо

Answer 1

1) x²<1
x²-1<0
(x-1)(x+1)<0
x принадлежит (-1;1)

2) (6x-7)(5x+1)/(x-3)≥0
х≠3
30(х-7/6)(х+1/5)/(х-3)≥0
7/6= 1 1/6
точки изменения знака 1 1/6, -1/5 и 3
х принадлежит [-1/5;1 1/6]U(3;+∞)

Answer 2

1) To solve the inequality x² < 1, we can start by factoring the expression x² - 1. This can be written as (x - 1)(x + 1). Now we can set each factor less than zero and solve for x.

Setting x - 1 < 0: x < 1

Setting x + 1 < 0: x < -1

Therefore, the solution to the inequality x² < 1 is -1 < x < 1. [[1]]

2) To solve the inequality (6x - 7)(5x + 1)/(x - 3) > 0, we can use the concept of sign charts.

First, we need to find the critical points of the expression (6x - 7)(5x + 1)/(x - 3). The critical points occur when the numerator or denominator is equal to zero.

Setting 6x - 7 = 0: x = 7/6

Setting 5x + 1 = 0: x = -1/5

Setting x - 3 = 0: x = 3

Now, we can create a sign chart by dividing the number line into intervals using the critical points. We will test a value from each interval to determine the sign of the expression.

Interval (-∞, -1/5): Choosing x = -1, we get: (6(-1) - 7)(5(-1) + 1)/(-1 - 3) = (-13)(-4)/(-4) = 13 > 0

Interval (-1/5, 7/6): Choosing x = 0, we get: (6(0) - 7)(5(0) + 1)/(0 - 3) = (-7)(1)/(-3) = 7/3 < 0

Interval (7/6, 3): Choosing x = 2, we get: (6(2) - 7)(5(2) + 1)/(2 - 3) = (5)(11)/(-1) = -55 < 0

Interval (3, ∞): Choosing x = 4, we get: (6(4) - 7)(5(4) + 1)/(4 - 3) = (17)(21)/(1) = 357 > 0

From the sign chart, we can see that the expression is positive in the intervals (-∞, -1/5) and (3, ∞). Therefore, the solution to the inequality (6x - 7)(5x + 1)/(x - 3) > 0 is x < -1/5 or x > 3. [[2]]