f(x)={x-1x＞0,0x=0,x+1x＜0}求f[f(3)]要过程

f(x)={x-1x＞0,0x=0,x+1x＜0}求f[f(3)]要过程

3>0
所以此时f(x)=x-1
所以f(3)=3-1=2
而2>1
所以f[f(3)]
=f(2)
=2-1
=1

原式=[(x+3)(x+9)][(x+5)(x+7)]+16
=[(x²+12x)+27][(x²+12x)+35]+16
=(x²+12x)²+62(x²+12x)+945+16
=(x²+12x)²+62(x²+12x)+961
=(x²+12x+31)²
所以是完全平方数