将x=1带入 z=√x+3+f(√y+1) 则z=4+ f(√y+1)= y²
则f (√y+1)= y²-4 则z=√x+3+ y²-4
设t=√y f(t+1)=t4-4(t的4次方)
设u=t+1,t=u-1 则f(u)=(u-1)4-4 ((u-1)的4次方)
将x=1带入 z=√x+3+f(√y+1) 则z=4+ f(√y+1)= y²
则f (√y+1)= y²-4 则z=√x+3+ y²-4
设t=√y f(t+1)=t4-4(t的4次方)
设u=t+1,t=u-1 则f(u)=(u-1)4-4 ((u-1)的4次方)