answer:To find the greatest common divisor (GCD) of the three numbers 638, 522, and 406, we can use the Euclidean algorithm, which involves a series of division steps to calculate the GCD. Let's start by finding the GCD of the first two numbers, 638 and 522: We divide 638 by 522: 638 = 522 times 1 + 116 Now, we use 522 and the remainder 116 to repeat the process: 522 = 116 times 4 + 74 Next, we divide 116 by the new remainder 74: 116 = 74 times 1 + 42 Continuing with 74 and 42: 74 = 42 times 1 + 32 Next, divide 42 by 32: 42 = 32 times 1 + 10 Now, divide 32 by 10: 32 = 10 times 3 + 2 Lastly, divide 10 by 2: 10 = 2 times 5 + 0 When we reach a remainder of 0, the divisor at this step is the GCD of the first two numbers. In this case, it is 2. Now we need to find the GCD of 2 and the third number 406. Using the Euclidean algorithm again: Divide 406 by 2: 406 = 2 times 203 + 0 Since there is no remainder, the divisor 2 is the GCD of 406 and 2. Therefore, the GCD of all three numbers, 638, 522, and 406, is boxed{2}.

answer:To find the intercepts a and b of the line x-2y-2=0 on the x-axis and y-axis respectively, we set the variables to 0 one at a time. For the y-axis intercept (b): - Let x=0 in the equation x-2y-2=0. - This gives us 0-2y-2=0. - Simplifying, we get -2y=2. - Dividing both sides by -2, we find y=-1. - Therefore, b=-1. For the x-axis intercept (a): - Let y=0 in the equation x-2y-2=0. - This gives us x-0-2=0. - Simplifying, we get x=2. - Therefore, a=2. Combining these results, we find that the intercepts are a=2 and b=-1. Therefore, the correct answer is boxed{text{B}}.

answer:**Analysis** Given the conditions, when "if a_1 < 0 and 0 < q < 1", it can be deduced that for any n in mathbb{N}^*, a_{n+1} > a_n. Therefore, it is a sufficient condition. Conversely, if the sequence is increasing, a_1 > 0 and q > 1 also hold, which means the conclusion cannot deduce the condition. Therefore, the correct choice is boxed{text{A}}.

answer:1. **Applying Ptolemy's Inequality:** We start by applying Ptolemy's inequality to the quadrilateral (ACDE). Ptolemy's inequality states that for any convex quadrilateral (ABCD), the following inequality holds: [ AC cdot BD + AD cdot BC geq AB cdot CD + BC cdot AD ] For the quadrilateral (ACDE), we have: [ AC cdot DE + CD cdot AE geq AD cdot CE ] Given that (CD = DE), we can rewrite this as: [ DE cdot (AC + AE) geq DA cdot CE ] Dividing both sides by (DA cdot (AC + AE)), we get: [ frac{DE}{DA} geq frac{CE}{AC + AE} ] 2. **Applying Ptolemy's Inequality to Other Quadrilaterals:** Similarly, we apply Ptolemy's inequality to the quadrilateral (CAEF): [ CA cdot EF + FA cdot CE geq CF cdot AE ] Given that (EF = FA), we can rewrite this as: [ FA cdot (CA + CE) geq FC cdot AE ] Dividing both sides by (FC cdot (CA + CE)), we get: [ frac{FA}{FC} geq frac{AE}{CA + CE} ] 3. **Applying Ptolemy's Inequality to Another Quadrilateral:** Finally, we apply Ptolemy's inequality to the quadrilateral (ABCE): [ AB cdot CE + BC cdot AE geq BE cdot AC ] Given that (AB = BC), we can rewrite this as: [ BC cdot (AE + AC) geq BE cdot AC ] Dividing both sides by (BE cdot (AE + AC)), we get: [ frac{BC}{BE} geq frac{AE}{AE + AC} ] 4. **Summing the Inequalities:** We now sum the three inequalities obtained: [ frac{DE}{DA} geq frac{CE}{AC + AE} ] [ frac{FA}{FC} geq frac{AE}{CA + CE} ] [ frac{BC}{BE} geq frac{AE}{AE + AC} ] Adding these inequalities, we get: [ frac{BC}{BE} + frac{DE}{DA} + frac{FA}{FC} geq frac{AE}{AE + AC} + frac{CE}{AC + AE} + frac{AE}{CA + CE} ] 5. **Applying Nesbitt's Inequality:** Nesbitt's inequality states that for positive real numbers (a, b, c): [ frac{a}{b+c} + frac{b}{c+a} + frac{c}{a+b} geq frac{3}{2} ] By comparing the sums, we see that: [ frac{AE}{AE + AC} + frac{CE}{AC + AE} + frac{AE}{CA + CE} geq frac{3}{2} ] Therefore, we conclude: [ frac{BC}{BE} + frac{DE}{DA} + frac{FA}{FC} geq frac{3}{2} ] (blacksquare)