分类：教育/科学-学习帮助
问题：初二数学综合计算题1.(a^2+3a+3/a^2+3a+2)
1.(a^2+3a+3/a^2+3a+2)-(a^2+3a-2/4-a^2)-(2a^2-2a-1/a^2-a-2)
回答：
(a^2+3a+3)/(a^2+3a+2)-(a^2+3a-2)/(4-a^2)
  -(2a^2-2a-1)/(a^2-a-2)

=[(a^2+3a+2)+1]/(a^2+3a+2)-[-(4-a^2)+3a+2]/(4-a^2)
  -[2(a^2-a-2)+1]/(a^2-a-2)

=[1+1/(a^2+3a+2)]-[-1-(3a+2)/(a^2-4)]-[2-1/(a^2-a-2)]

=1/[(a+1)(a+2)]+(3a+2)/[(a-2)(a+2)]+1/[(a-2)(a+1)]

=[(a-2)+(3a+2)(a+1)+(a+2)]/[(a+1)(a-2)(a+2)]

=(3a^2+7a+2)/[(a+1)(a-2)(a+2)]
 
=(3a+1)(a+2)/[(a+1)(a-2)(a+2)]

=(3a+1)/[(a+1)(a-1)].