9 8 7 6 5 4 3 2 1=2001,在空格处添加符号,即加减乘除以及括号.

987+6+(5+43)*21=2001,(9-8+7*6)*54-321=2001
9*87*(6/54*32-1)=2001,9*87+(65-4-3)*21=2001
(9*87+654/3)*2-1=2001,98/7*6*5*4+321=2001
(98-7)*(65-43)-2+1=2001,98*7+(654+3)*2+1=2001
(98/7+654)*3-2-1=2001,(98-76)*(5+43*2)-1=2001
(98-765)*(4-3*2-1)=2001,(98-765)/(4/3/2-1)=2001
(987-6*5+43)*2+1=2001,9-8*((7+6+5)*4-321)=2001
9*8*7/6*5*4+321=2001,9+8*(7+6+5*43+21)=2001
9+8*(7+(6+5)*(43-21))=2001,9-8+(7*6+5)*43-21=2001
((9+8)*7*6-54)*3+21=2001,9+8*((7+65+4)*3+21)=2001
9+8*(7+65*4+3-21)=2001,(9+8/7*65+4*3)*21=2001
(9-(8/7-65*4)/3)*21=2001,9+8-(7-65-4)*32*1=2001
(9+8-(7-65-4)*32)*1=2001,9+(8-(7-65-4)*32)*1=2001
9+8-(7-65-4)*32/1=2001,(9+8-(7-65-4)*32)/1=2001
9+(8-(7-65-4)*32)/1=2001,9-(8*(7-65*4)+32)*1=2001
9-(8*(7-65*4)+32)/1=2001,9-8*(7-65*4)-32*1=2001
(9-8*(7-65*4)-32)*1=2001,9-8*(7-65*4)-32/1=2001
(9-8*(7-65*4)-32)/1=2001,9+8*(7*65+43)/2*1=2001
(9+8*(7*65+43)/2)*1=2001,9+8*(7*65+43)/2/1=2001
(9+8*(7*65+43)/2)/1=2001,9+8+(7+654)*3+2-1=2001
(9+8-7+654+3)*(2+1)=2001,9+(8+76-5+4)*(3+21)=2001
9+8*(76*(5+4)/3+21)=2001,9+8*(76*(5-4)*3+21)=2001
9+8*(76/(5-4)*3+21)=2001,9+8*((76+5+43)*2+1)=2001
(9+8*(76+5+43))*2-1=2001,9-8*(76-54*3*2-1)=2001
9+(87-(6-5)*4)*(3+21)=2001,9+(87*(6-5)-4)*(3+21)=2001
9+(87/(6-5)-4)*(3+21)=2001,9*87*((6/54+3)/2+1)=2001
(9+87*(65+4))/3-2-1=2001,(98-7+6*5/(4+3))*21=2001
(98*7-6-5*4)*3+21=2001,(98*7-6*5+4)*3+21=2001
(98-7)*(6*(5+4)-32)-1=2001,(98-7)*(6-5+43)/2-1=2001
(98+7+6)*54/3+2+1=2001,98-(7-6*54)*3*2+1=2001
(98-7)*((6+54)/3+2)-1=2001,(98-7)*(6+54/3-2)-1=2001
(98+76)* ...展开全部>
lucifer0728 | 2011-04-22
18
12987654321中间都有空的。。。要每个数字间都有符号而且数字顺序不变(9*(8+7*6*5+4)+3)*(2-1)=2001

9*(8+7*6*5+4)+3/(2-1)=2001

9*(8+7*6*5+4)+3*(2-1)=2001非常感谢