Question

Помогите пожалуйста 1) (x-1)(2x+5)/x-3>0 2) 7+x/(4-x)(2x-3)<0 3) (x+1)^2(x+5)/x-3>0 4)x-4/(4-x)(2x-3)<0 5)x^2-x-20/x+4>0 Срочно!!!

Answer 1

ВАРИАНТ 9 1) 15х2 + 4 = 16х 2) 7х2 = 4х – 3 3) 2х – 5х2 = 0 4) 5х2 – 20 = 0 5) 7х + 3 + 4х2 = 0 6) х2 – 9х + 18 = 0 7) 16k2 + 9 – 24k = 0

Answer 2

1) To solve the inequality (x-1)(2x+5)/(x-3) > 0, we can use the sign chart method. First, we need to find the critical points by setting the numerator and denominator equal to zero. The critical points are x = 1, x = -5/2, and x = 3. We can then create a sign chart and test the intervals between these critical points.

- When x < -5/2, both the numerator and denominator are negative, so the expression is positive. - When -5/2 < x < 1, the numerator is negative and the denominator is positive, so the expression is negative. - When 1 < x < 3, both the numerator and denominator are positive, so the expression is positive. - When x > 3, the numerator is positive and the denominator is positive, so the expression is positive.

Therefore, the solution to the inequality is x < -5/2 or 1 < x < 3. (Answer: x < -5/2 or 1 < x < 3). [[1]]

2) To solve the inequality 7 + x / ((4 - x)(2x - 3)) < 0, we can follow a similar approach. First, we find the critical points by setting the denominator equal to zero. The critical points are x = 4 and x = 3/2. We can then create a sign chart and test the intervals between these critical points.

- When x < 3/2, both the numerator and denominator are negative, so the expression is positive. - When 3/2 < x < 4, the numerator is positive and the denominator is negative, so the expression is negative. - When x > 4, both the numerator and denominator are positive, so the expression is positive.

Therefore, the solution to the inequality is x < 3/2 or x > 4. (Answer: x < 3/2 or x > 4). [[2]]

3) To solve the inequality (x+1)^2(x+5)/(x-3) > 0, we can again use the sign chart method. The critical points are x = -5, x = -1, and x = 3. We can create a sign chart and test the intervals between these critical points.

- When x < -5, both the numerator and denominator are negative, so the expression is positive. - When -5 < x < -1, the numerator is positive and the denominator is negative, so the expression is negative. - When -1 < x < 3, both the numerator and denominator are positive, so the expression is positive. - When x > 3, the numerator is positive and the denominator is positive, so the expression is positive.

Therefore, the solution to the inequality is x < -5 or -1 < x < 3. (Answer: x < -5 or -1 < x < 3). [[3]]

4) To solve the inequality (x-4)/((4-x)(2x-3)) < 0, we can once again use the sign chart method. The critical points are x = 4 and x = 3/2. We can create a sign chart and test the intervals between these critical points.

- When x < 3/2, both the numerator and denominator are negative, so the expression is positive. - When 3/2 < x < 4, the numerator is negative and the denominator is positive, so the expression is negative. - When x > 4, both the numerator and denominator are positive, so the expression is positive.

Therefore, the solution to the inequality is x < 3/2 or x > 4. (Answer: x < 3/2 or x > 4). [[4]]

5) To solve the inequality (x^2 - x - 20)/(x + 4) > 0, we can factor the numerator as (x - 5)(x + 4) and rewrite the inequality as (x - 5)(x + 4)/(x + 4) > 0. We can then cancel out the common factor of (x + 4) and simplify the inequality to (x - 5) > 0.

Therefore, the solution to the inequality is x > 5. (Answer: x > 5). [[5]]

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