解：3(a+1)²=-5(a+1)(a-1)+2(a-1)²
=[3(a+1)+2(a-1)][(a+1)+(a-1)]
=2a(5a+1)
当a=-2分之1时
原式=2×(-2分之1)×[5×(-2分之1)+1]
=-(-2分之5+1)
=-(-2分之3)
=2分之3
3(a+1)^2-5(a+1)(a-1)+2(a-1)^2
=3a^2+6a+3-5a^2+5+2a^2-4a+2
=2a+10
当a=-1/2时，
原式=-1+10=9
3(a+1)^2-5(a+1)(a-1)+2(a-1)^2
=[(a+1)-(a-1)][(3(a+1)-2(a-1)]
=(a+1-a+1)(3a+3-2a+2)
=2(a+5)
=2×(-1/2+5)
=2×9/2
=9
内容全部来源于网络收集，如有侵权，请联系网站删除：QQ：24596024