设t=√(x+y+z+5) (t≥0,且x+y+z≥-5)
则,x+y+z=t²-5
则:2(t²-5)-5t-2=0——>2t²-5t-12=0——>(2t+3)(t-4)=0——>t=-3/2或t=4
显然,t=-3/2不符合t≥0
∴t=4即:√(x+y+z+5)=4——>x+y+z+5=16——>x+y+z=11
又∵x/3=y/4=z/5
∴(x+Y+z)/(3+4+5)=x/3=y/4=z/5 (根据等比关系定理)
∴x/3=y/4=z/5=11/12
∴x=11/4,y=11/3,z=55/12