(60-50+x)(800-100x)=12000具体过程

问题描述:

(60-50+x)(800-100x)=12000具体过程

(60-50+x)(800-100x)=12000
48000-40000+800x-6000x+5000x=12000
-200X=4000
x=-20

8000-1000x+800x-100x²=12000即-100x²-200x-4000=0也就是x²+2x+40=0
(x+1)²=-41所以x=[41^(1/2)]i-1
这涉及到复数运算。i²=-1

(60-50+x)(800-100x)=12000
(10-x)(8-x)×100=12000
(10-x)(8-x)=120
80-8x-10x+x²=120
x²-18x-40=0
(x-20)(x+2)=0
x-20=0 或者 x+2=0
x=20 或者x=-2