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方程(x+3)(2x-5)-(2x+1)(x-8)=41的解是   .

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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 cl[]='1','2','3','4','5';  char c2[]='/x10','/xa','/x8';  
x6÷x3=x2B  (x-1)2=x2-1  x4+x4=x8  (x-1)2=x2-2x+1  
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'};  
x2·x3=x6   (x2)3=x8   x2+x3=x5   x6÷x3=x3  
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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';  
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'};