如图数表: ⎜⎜⎜a11a21…an1a12a22…an2………a1na2n…ann⎞⎠⎟⎟⎟,每一行都是首项为1的等差数列,第m行的公差为dm,且每一列也是等差数列,设第m行的第k项为amk(m,k=1,2,3,…,n,n⩾3,n∈N∗). (1)证明:d1,d2,d3成等差数列,并用m,d1,d2表示dm(3⩽m⩽...
设a1,a2,⋯, a11为正整数,其中a1<a2<⋯<a11.若1a1,1a2,⋯,1a11成一等差数列(即 1a2−1a1=1a3−1a2=⋯=1a11−1a10 ),求a1的最小可能值. 相关知识点: 试题来源: 解析2310.Let M be the L.C.M. of a1, a2, ⋯,
型号 MSDA5A5A1A 价格说明 价格:商品在爱采购的展示标价,具体的成交价格可能因商品参加活动等情况发生变化,也可能随着购买数量不同或所选规格不同而发生变化,如用户与商家线下达成协议,以线下协议的结算价格为准,如用户在爱采购上完成线上购买,则最终以订单结算页价格为准。 抢购价:商品参与营销活动的活动价格,...
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ann Aij表示元素aij的代数余子式 则下列公式成立 ak1Ai1 ak2Ai2 aknAin 25用连加号简写为n 武汉大学第1章行列式April14 201451 161设D a11a12 a1na21a22 a2n an1an2 ann Aij表示元素aij的代数余子式 则下列公式成立 ak1Ai1 ak2Ai2 aknAin 25用连加号简写为n 武汉大学第1章行列式April14 201451 ...
3.解: P-Q=(1/(a_1)+1/(a_2))-(1/(a_1a_2)+1)= a1 a2 a1a2 (a_2+a_1)/(a_1a_2)-(1+a_1a_2)/(a_1a_2)=(a_1-1+a_2(1-a_1)/( =((a_1-1)(1-a_2))/(a_1a_2) 因为 a_11 ,a21,所以 a_1-10 ,1-a2 0,a_1a_20 , 所以 P-Q=((a_1-1...
结果1 题目7.按第3列展开下列行列式,并计算其值:0a1a11 a12 a13 a14 a151b-1a21 a22 a233 a24 a250-1(1)100-1-1 c-1解答 (2)a31 a32-10a41 a42001a51 a52000 相关知识点: 试题来源: 解析 答案(小) (2)0 解析 l) -( -( -( o E , -b⋅(-1)^(1+3)*|*(0-1)+c(-1)^...
a1coho1 drinking pattem and a11 cause cardiovascu1ar and a1coho1 re1ated morta1ity in eastem europeAlcohol has been implicated in the high mor- tality in Central and Eastern Europe but the magnitude of its effect, and whether it is due to regular high intake or episodic binge drinking ...
embo1ization of a rena1 so1id b1eeding 1esion in in10sive care unit in a critica11y i11 patient with severe acute respiratory syndrome re1ated coronavirus infection and difficu1t mobi1ization due to extracorporea1 membrane oxygenation and continuous rena1 rep1acement therapy ...
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