zhijunwin
幼苗
共回答了24个问题采纳率:87.5% 举报
1. 设y=(5x+2)/ (x+3)
则x=(3y-2)/(5-y)>-2
(3y-2)/(5-y)+2>0
[(3y-2)+2(5-y)]/(5-y) >0
-8<y<5
即:-8<(5x+2)/ (x+3)<5
2.原式=√[(a-1)+2√(a-1)+1]+√[(a-1)-2√(a-1)+1] =[√(a-1)+1]+∣√(a-1)-1∣
当a>2时 =2√(a-1)
当1≤a≤2时 =2
3.用待定系数法:
设(x-1)(x^2+ax+b)=x^3+(a-1)x^2+(b-a)x-b=x^3-kx^2+x+1
得出:a-1=-k,b-a=1,-b=1,解出:k=3,a=-2,b=-1
(x-1)(x^2-2x-1)=(x-1) (x-1-√2)(x-1+√2)
1年前
6