13问答网 > a^2+a-2⼀a^2-2a-8+a^2-4a-21⼀a^2-a-12-a^2+3a-18⼀a^2+6a

a^2+a-2⼀a^2-2a-8+a^2-4a-21⼀a^2-a-12-a^2+3a-18⼀a^2+6a

是用配方法的吗

2025-03-03 10:53:54

推荐回答（3个）

回答1：

a^2+a-2/a^2-2a-8+a^2-4a-21/a^2-a-12-a^2+3a-18/a^2+6a
=(a+2)(a-1)/(a-4)(a+2)+(a-7)(a+3)/(a-4)(a+3)-(a+6)(a-3)/a(a+6)
=(a-1)/(a-4)+(a-7)/(a-4)-(a-3)/a
=(2a-8)/(a-4)-(a-3)/a
=2-1+3/a
=1+(3/a)

回答2：

a^2+a-2/a^2-2a-8+a^2-4a-21/a^2-a-12-a^2+3a-18/a^2+6a
=(a+2)(a-1)/(a-4)(a+2)+(a-7)(a+3)/(a-4)(a+3)-(a+6)(a-3)/a(a+6)
=(a-1)/(a-4)+(a-7)/(a-4)-(a-3)/a
=(a-1+a-7)/(a-4)-(a-3)/a
=2-(a-3)/a
=(2a-a+3)/a
=(a+3)/a

回答3：

LZ是要合并同类项么如果是的话结果为a^2+7a-41/a^2-20