answer:1. **Define the Functions:** We start by defining the functions on either side of the given inequality: [ varphi(e) = left(a e + frac{b}{e}right)left(c e + frac{d}{e}right) ] [ psi(f) = left(a^2 f^2 - frac{b^2}{f^2}right)left(frac{d^2}{f^2} - c^2 f^2right) ] 2. **Expand the Function varphi(e):** [ varphi(e) = left(a e + frac{b}{e}right) left(c e + frac{d}{e}right) ] Expanding this, we get: [ varphi(e) = a c e^2 + a d + frac{b c}{e^2} + frac{b d}{e^2} ] 3. **Applying the Cauchy-Schwarz Inequality:** By the Cauchy-Schwarz inequality, we have: [ a c e^2 + frac{b d}{e^2} geq 2 sqrt{a c b d} ] Hence: [ a c e^2 + a d + frac{b c}{e^2} + frac{b d}{e^2} geq a d + b c + 2 sqrt{a c b d} ] Therefore, we can write: [ varphi(e) = a c e^2 + a d + frac{b c}{e^2} + frac{b d}{e^2} geq (sqrt{a d} + sqrt{b c})^2 ] 4. **Expand the Function psi(f):** [ psi(f) = left(a^2 f^2 - frac{b^2}{f^2}right) left(frac{d^2}{f^2} - c^2 f^2right) ] Expanding this, we get: [ psi(f) = a^2 d^2 - a^2 c^2 f^4 - frac{b^2 d^2}{f^4} + b^2 c^2 ] 5. **Applying the Rearrangement Inequality:** By the rearrangement inequality: [ psi(f) = a^2 d^2 + b^2 c^2 - left(a^2 c^2 f^4 + frac{b^2 d^2}{f^4}right) leq a^2 d^2 + b^2 c^2 - 2 a c b d ] Therefore, we can write: [ psi(f) leq (a d - b c)^2 ] 6. **Combine Both Inequalities:** For any (e, f > 0), we see: [ psi(f) leq (sqrt{a d} - sqrt{b c})^2 (sqrt{a d} + sqrt{b c})^2 ] Rewriting: [ (sqrt{a d} - sqrt{b c})^2 (sqrt{a d} + sqrt{b c})^2 = ((sqrt{a d})^2 - (sqrt{b c})^2)^2 = (ad - bc)^2 ] 7. **Final Step:** Since (left| sqrt{ad} - sqrt{bc} right| leq 1), substituting the value will always maintain: [ psi(f) leq (sqrt{ad} - sqrt{bc})^2 leq (sqrt{ad} + sqrt{bc})^2 leq varphi(e) ] Therefore, the final inequality holds: [ varphi(e) geq psi(f) ] Hence, we conclude: [ boxed{left(a e + frac{b}{e}right)left(c e + frac{d}{e}right) geq left(a^2 f^2 - frac{b^2}{f^2}right)left(frac{d^2}{f^2} - c^2 f^2right)} ]

answer:To find the value of x that satisfies all the given conditions, we need to find the intersection of all the ranges provided. 1. (1 < x < 9) 2. (2 < x < 15) 3. (7 > x > -1) 4. (4 > x > 0) 5. (x + 1 < 5) Let's analyze each condition: 1. (x) must be greater than 1 and less than 9. 2. (x) must be greater than 2 and less than 15. This narrows down the range from the first condition to (2 < x < 9). 3. (x) must be less than 7 and greater than -1. This does not further narrow down the range since (2 < x < 9) is already within (7 > x > -1). 4. (x) must be less than 4 and greater than 0. This further narrows down the range to (2 < x < 4). 5. (x + 1 < 5) simplifies to (x < 4). This does not change the range since (2 < x < 4) is already within (x < 4). So the final range for (x) that satisfies all conditions is (2 < x < 4). Since (x) is an integer, the only integer values that satisfy this range are 3. Therefore, (x = boxed{3)} .

answer:To solve the given problem, we follow these steps: 1. **Solve |x-1| < 4:** Given |x-1| < 4, we can break this into two cases based on the definition of absolute value: - Case 1: x-1 < 4 - Adding 1 to both sides, we get x < 5. - Case 2: -(x-1) < 4, which simplifies to x-1 > -4 - Adding 1 to both sides, we get x > -3. Combining both cases, we find the solution set A = {x | -3 < x < 5}. 2. **Solve frac{x-5}{2-x} > 0:** For the inequality frac{x-5}{2-x} > 0, we identify the critical points by setting the numerator and denominator to zero: - Numerator x-5 = 0 gives x = 5. - Denominator 2-x = 0 gives x = 2. The critical points divide the real number line into intervals. We test each interval in the inequality: - For x < 2, choose x = 1, we get frac{1-5}{2-1} = -4 < 0. - For 2 < x < 5, choose x = 3, we get frac{3-5}{2-3} = 2 > 0. - For x > 5, choose x = 6, we get frac{6-5}{2-6} = -frac{1}{4} < 0. Thus, the solution set is B = {x | 2 < x < 5}. 3. **Compare Sets A and B:** Given A = {x | -3 < x < 5} and B = {x | 2 < x < 5}, we observe that every element in B is also in A, but not every element in A is in B. This means B is a subset of A (B subset A), indicating that the condition |x-1| < 4 (set A) is a necessary but not sufficient condition for frac{x-5}{2-x} > 0 (set B). Therefore, the correct answer is: boxed{text{B: necessary but not sufficient condition}}

answer:1. **Given Information**: The pentagon (ABCDE) is inscribed in the circle (omega) and diagonal (AC) is the diameter of (omega). We are also given that (angle ADB = 20^circ). 2. **Arch Interpretation**: Since (angle ADB = 20^circ), the arc (AB) subtended by this angle is twice this angle, because the angle subtended by the arc at the circumference is half of the angle at the center. Thus, the arc (AB) is: [ text{Arc } AB = 2 times 20^circ = 40^circ ] 3. **Arc Calculation**: Since (AC) is the diameter of the circle, the total central angle subtended by (AC) is (180^circ). The arc (AB) is part of this (180^circ), so the arc (BC) is: [ text{Arc } BC = 180^circ - 40^circ = 140^circ ] 4. **Determining (angle BEC)**: The angle (angle BEC) subtended by the arc (BC) at the circumference of the circle is half of the central angle subtended by the same arc. Therefore, [ angle BEC = frac{text{Arc } BC}{2} = frac{140^circ}{2} = 70^circ ] By following these steps, we have determined that: [ boxed{70^circ} ]