已知(2a加b加3)的平方加【b减1】等于0,求5a加{负2a减3【2b减8加(3a减2b减1)减a】加1}的值
问题描述:
已知(2a加b加3)的平方加【b减1】等于0,求5a加{负2a减3【2b减8加(3a减2b减1)减a】加1}的值
答
﹙2a+b+3﹚²+﹙b-1﹚=0求5a+﹛-2a-3[2b-8+﹙3a-2b-1﹚-a]+1﹜=?
2a+b+3=0且b-1=0 得b=1 a=-2
5a+﹛-2a-3[2b-8+﹙3a-2b-1﹚-a]+1﹜
=5a+﹛-2a-3[2b-8+3a-2b-1-a]+1﹜
=5a+[-2a-3﹙2a-9﹚+1]
=5a+﹙-2a-6a+27+1﹚
=5a-8a+28
=-3a+28
=6+28
=34
答
﹙2a+b+3﹚²+﹙b-1﹚=0求5a+﹛-2a-3[2b-8+﹙3a-2b-1﹚-a]+1﹜=?
2a+b+3=0且b-1=0 得b=1 a=-2
5a+﹛-2a-3[2b-8+﹙3a-2b-1﹚-a]+1﹜
=5a+﹛-2a-3[2b-8+3a-2b-1-a]+1﹜
=5a+[-2a-3﹙2a-9﹚+1]
=5a+﹙-2a-6a+27+1﹚
=5a-8a+28
=-3a+28
=6+28=34
答
(2a加b加3)的平方加【b减1】等于0,
2a+b+3=0
b-1=0 所以b=1 a=-2
5a加{负2a减3【2b减8加(3a减2b减1)减a】加1}
=-10+{6-3[2-8+(-6-2-1)+2]+1}
=36