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1.已知:xy+x+y=71,x~2y+xy~2=880,x,y为正整数,求x~2十y~2. 2.矩形ABCD,AB=4,CB=3,点A=p_O,P_1,…,P_(168)=B把AB边分为168个相等的小段,点C=Q_o,Q_1,…,Q(168)=B把CB边分成168个相等的小段,做线段P_kQ_K,1≤k≤167,在AD,CD上同样重复上述过程,再引对角线AC,求这335条线段长度之和。 3.把(1+0.2)~(1000)按二项式定理展开,且令c_1000~0(0.2)~0+C_1000~1(0.2)~1+…+C_1000~k(0.2)~1000=A_o+A_1+…+A_1000,其中A_=C_1000~1000(0.2)~k,k=0,1,2,…,1000,问使A_k最大时k是多少?
1. Known: xy + x + y = 71, x ~ 2y + xy ~ 2 = 880, x, y is a positive integer, seeking x ~ 2 ten y ~ 2. 2. Rectangular ABCD, AB = 4, CB = 3. Points A=p_O, P_1,..., P_(168)=B divide the AB edge into 168 equal segments, and points C=Q_o, Q_1,..., Q(168)=B divide the CB edge into 168 equal segments. In the small segment, do the line segment P_kQ_K, 1≤k≤167, repeat the same process on AD, CD, and then lead the diagonal AC to find the sum of the lengths of the 335 segments. 3. Expand (1+0.2)~(1000) according to the binomial theorem and let c_1000~0(0.2)~0+C_1000~1(0.2)~1+...+C_1000~k(0.2)~1000= A_o+A_1+...+A_1000, where A_=C_1000~1000(0.2)~k,k=0,1,2,...,1000. What is the value of k when A_k is maximized?