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设(+x)10=a0+a1x+a2x2+…+a10x10,则a2=   ,(a0+a2+a4+…+a10)2﹣(a1+a3+a5+…+a9)2的值为   .

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double x[5]=2.0, 4.0, 6.0, 8.0, 10.0;  int y[5]=0, 1, 3, 5, 7, 9;  char c1[]='1', '2', '3', '4', '5';  char c2[]='/x10', '/xa', '/x8';  
{x|x≤1}  {x|1≤x<2}  {x|0<x≤1}  {x|0<x<1}  
double x[5] =2.0,4.0,6.0,8.0,10.0;  int y[5] =0,1,3,5,7,9;  char c 1[ ] ='1','2','3','4','5';  char c 2 [C] =,'/x10','/xa','/x8';  
double x[5]=2.0,4.0,6.0,8.0,10.0;  int y[5]=0,1,3,5,7,9;  char cl[]='1','2','3','4','5';  char c2[]='/x10','/xa','/x8';  
double x[5]=2.0,4.0,6.0,8.0,10.0;  int y[5.3]=0,1,3,5,7,9;  charc/[]='1','2','3','4','5';  char c2[]='/x10','/xa','/x8';  
double x[5]={2.0,4.0,6.0,8.0,10.0};  int y[5]={0,1,3,5,7,9};  char c1[]={'1','2','3','4','5'};  char c2[]={'/x10','/xa','/x8'};  
(-2,2)  (1,2)  -1,0,1  -2,-1,0,1,2  
(-∞,0)∪(10,+∞)   (-1,+∞)   (-∞,-2)∪(-1,10)   (0,10)  
double x[5]=2.0,4.0,6.0,8.0,10.0;  int y[5]=0,1,3,5,7,9;  char c1[]=‘1’,’2’,’3’,’4’,’5’;  char c2[]=‘/x10’,’/xa’,’/x8’;  
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double x[5] ={2.0,4.0,6.0,8.0,10.0};  int y[5] ={0,1,3,5,7,9};  char c 1[ ] ={'1','2','3','4','5'};  char c 2 ={,'/x10','/xa','/x8'};  

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