Question

(x-1)(x-4)(√3+5x-2x^2)<0

Answer 1

Ответ:

Объяснение: данное неравенство равносильно системе двух неравенств: {(x-1)(x-4)<0,3+5x-2x²>0

{(x-1)(x-4)<0,2x²-5x-3<0

{(x-1)(x-4)<0, 2(x-3)(x+1/2)<0--- решаем нер-ва методом интервалов и находим  пересече  ние их решений:

----- (1)------------ (4)----------⇒  

+       ↓      -    ↑             +           }    ⇒пересечение решений : х∈(1;3).

-(-1/2)---------(3)----------------⇒

+           -                 +    

Answer 2

To solve the inequality (x-1)(x-4)(√3+5x-2x^2) < 0, we can break it down into three separate cases and analyze the sign of each factor. The three cases are:

Case 1: (x-1) > 0, (x-4) > 0, (√3+5x-2x^2) < 0 Case 2: (x-1) < 0, (x-4) < 0, (√3+5x-2x^2) > 0 Case 3: (x-1) < 0, (x-4) > 0, (√3+5x-2x^2) < 0

Let's solve each case separately:

Case 1: (x-1) > 0, (x-4) > 0, (√3+5x-2x^2) < 0

For (x-1) > 0, x > 1 For (x-4) > 0, x > 4

To solve (√3+5x-2x^2) < 0, we can first find the critical points by setting (√3+5x-2x^2) = 0 and solving for x.

√3 + 5x - 2x^2 = 0 2x^2 - 5x - √3 = 0

Using the quadratic formula, we have:

x = (5 ± √(5^2 - 4(2)(-√3))) / (2(2)) x = (5 ± √(25 + 8√3)) / 4 x ≈ 2.04 or x ≈ 0.46

Now, we can analyze the sign of (√3+5x-2x^2) for the intervals defined by these critical points and the inequalities x > 1 and x > 4.

For x < 0.46, (√3+5x-2x^2) < 0 For 0.46 < x < 2.04, (√3+5x-2x^2) > 0 For x > 2.04, (√3+5x-2x^2) < 0

Therefore, for Case 1, the solution is 1 < x < 2.04.

Case 2: (x-1) < 0, (x-4) < 0, (√3+5x-2x^2) > 0

For (x-1) < 0, x < 1 For (x-4) < 0, x < 4

Since (√3+5x-2x^2) > 0 for this case, there are no solutions in this range.

Case 3: (x-1) < 0, (x-4) > 0, (√3+5x-2x^2) < 0

For (x-1) < 0, x < 1 For (x-4) > 0, x > 4

Since (√3+5x-2x^2) < 0 for this case, there are no solutions in this range.

Therefore, the solution to the inequality (x-1)(x-4)(√3+5x-2x^2) < 0 is 1 < x < 2.04.