函数y=(x^2+5)(x^2+1)的植域
热心网友
y=(x^2+5)(x^2+1)=x^4+6x^2+5=(x^2+3)^2-4∵x^2+3=3==(x^2+3)^2=9∴y=(x^2+3)^2-4=9-4=5∴y∈[5,∞)
热心网友
设X*X=T,Y=(T+5)/(T+1)=1+4/(T+1)因为T=0,所以4/(T+1)<=4,所以Y<=5.
热心网友
你改我也改:y=(x^2+5)/(x^2+1)y(x^2+1)=(x^2+5)(y-1)x^2=5-yx^2=(5-y)/(y-1)∵x^2≥0∴(5-y)/(y-1)≥0∴(5-y))≥0y-1)＞0∴1＜y≤5或(5-y)≤0y-1＜0(此不等式组无解)所以函数值域为1＜y≤5
热心网友
令t=x^2,则y=(t+5)(t+1)且t=0,所以y=5
热心网友
(x^2+5)(x^2+1)＝x^4＋6×x^2＋5＝（x^2＋3）^2－4≥3^2－4＝9-4＝5等号仅在x＝0时成立∴y∈[5，+∞)
热心网友
x^2=0,x^2+5=5,x^2+1=1y=5*1=5
热心网友
y=5